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TEXT  BOOK   OF  MECHANICS 

Designed  for  Colleges  and  Technical  Schools 
By  LOUIS   A.    MARTIN,   Jr. 

PUBLISHED    BY 

JOHN  WILEY    &    SONS,  Inc. 


Vol.    I.  Statics,      xii +147  pages,  5  by  7':^,  167  figures, 
Cloth,  $1.25  net. 

Vol.  FI.  Krnematics    and     Kinetics.        xiv+  223 

pages,   5  by  73:4,  91  figures.      Cloth,  $1.50  net. 

Vol.   III.   Meclianics  of  Materials,  .xili +229  pages, 
5  by  7'4,  III  figures.      Cloth,  $1.50  net. 

Vol.   IV.   Applied  Statics,     xii +198  pages,  5  by  7I4, 
141   figures.      Cloth,   $1.50  net. 

Vol.  V.   Hydraulics.      xii  +  223    pages,  5  by  73^,  114 
figures.      Cloth,  $1.50  net. 

Vol.    VI.  Thermodynamics,     xviii+313    pages,    5 
by  734,  78  figures.      Cloth,  $1.75  net.  j 


TEXT-BOOK 


OF 


ECHANICS 


BY 

LOUIS    A.    MARTIN,    Jr 

Professor  of  Mechanics 
Stevens  Instilute  of  Technology 


Vol.  VI. 

THERMODYNAMICS 


FIRST    EDITION 

FIRST    THOUSAND 


NEW  YORK 

JOHN  WILEY  &  SONS,   Inc. 

London:    CHAPMAN   &   HALL,  Limited 

1916 


Copyright,   1916 

BY 

LOUIS   A.   MARTIN,  Jr. 


THE  SCIENTIFIC  PRESS 

ROBERT   DRUMMOND  AND   COMPA^ 

BROOKLYN,  N.  Y. 


PREFACE 


The  kind  reception  of  the  first  five  volumes  of  this  text- 
book by  both  teachers  and  students  has  led  me  to  continue 
the  same  method  of  presentation. 

I  again  seek  to  produce  a  text  which  will  encourage  the 
student  to  think  and  not  to  memorize,  to  do  and  not  simply 
to  accept  something  already  done  for  him.  At  the  same 
time  care  has  been  taken  to  furnish  sufl&cient  material  in 
the  way  of  explanation  and  example  so  that  the  student 
may  not  become  discouraged. 

In  solving  examples  and  exercises  relating  to  steami  the 
Steam  Tables  and  Diagrams  of  Marks  and  Davis  (Long- 
mans, Green,  and  Co.)  have  been  used. 

I  hereby  gratefully  acknowledge  the  assistance  of  my  wife, 
Alwynne  B.  Martin,  both  in  the  preparation  of  the  manu- 
script and  in  the  reading  of  the  proof.  Mr.  Gustav  G. 
Freygang  has  also  been  kind  enuf  to  read  both  manu- 
script and  proof. 

L.  A.  M.,  Jr. 

Castle  Point,  Hoboken,  N.  J. 
January,  1016 


CONTENTS 


Introduction 

PAGE 

Thermodynamics ^ 

Energy ^ 

Heat  Is  a  Form  of  Energy i 

Internal  Kinetic  Energy 2 

Internal  Potential  Energy 2 

External  Work 3 

Conservation  of  Energy 3 

The  First  Law  of  Thermodynamics S 

Thermal  Capacity 5 

Specific  Heat 6 

GASES 

CHAPTER  I 

The  Laws  of  Ideal  Gases 

section  i 

Joule's,  Boyle's,  and  Charles'  Law 

Definitions 9 

Joule's  Law ^° 

Boyle's  Law i  ^ 

Law  of  Charles ^^ 

Absolute  Zero i  - 

The  General  Law  for  Ideal  Gases 13 

Exercise  i ^S 

Exercises  2  to  4 ^o 


VI  CONTENTS 

SECTION  II 
The  Gas  Constant  of  Any  Ideal  Gas 

PAGE 

Avogadro's  Law i6 

Exercises  5  to  g , 17 

SECTION   III 

The  Gas  Constant  of  Gaseous  Mixtures 

Dalton's  Law 18 

Exercises  10  to  14 19 

SECTION    IV 

The  Specific  Heat  of  Gases 

The  Specific  Heat  at  Constant  Volume 21 

Exercises  15  and  16 23 

The  Specific  Heat  at  Constant  Pressure 23 

The  Relation  between  Cp  and  c^ 25 

Specific  Heats  at  Constant  Pressure 26 

Exercises  ly  to  25 27 

SECTION    V 

The  Fundamental  Lines  for  Ideal  Gases 

The  Fundamental  Laws 28 

Exercise  26 29 

CHAPTER  H 

Changes  or  State  of  Ideal  Gases 

SECTION    VI 

General  Discussion 

Characteristic  Equation  and  Surface 30 

Exercises  27  and  28 32 

Important  Changes  of  State 2>2> 


CONTENTS  V)l 

SECTION  VII 
Isothermal,  Isometric,  Isopicstic,  aid  Isodynamic  Changes 

PAGE 

Isothermal  Change  of  State 34 

Exercises  2g  to  31 34 

Exercise  32 35 

Exercises  33  to  35 36 

Isometric  Change  of  State 36 

Exercises  36  to  39 36 

Isopiestic  Change  of  State 37 

Exercises  40  to  43. 37 

Isodynamic  Change  of  State 37 

Exercise  44 38 

SECTION  via 
Adiabatic  Change  of  State  of  Ideal  Gases 

Definition  and  General  Discus.sion 38 

Exercises  45  and  46 -39 

The  Equations  Governing  Adiabatic  Change  of  Stale 39 

Exercises  47  to  4g 41 

Exercises  50  and  51 42 

Exponential  and  Logarithmic  Computations 42 

Exercises  52  and  S3 43 

The  External  Work  Performed 44 

Exercises  54  to  58 44 

SECTION  IX 

Polytropic  Change  of  Slate  of  Ideal  Gases 

Definition  and  General  Discussion 45 

The  Equations  Governing  Poljlropic  Change 46 

Exercises  59  to  61 48 

Discussion  of  the  Polytropic  Equations 49 

Exercises  62  to  64 52 

The  Signs  of  dQ,  dK,  and  dW 53 

Exercise  65 55 

The  Experimental  Determination  of  » 56 


Vlll  '  CONTENTS 

PAGE 

Exercises  66  to  68 57 

The  External  Work  Performed 58 

Exercises  6g  to  71 58 

The  Heat  SuppHed 59 

Exercises  72  to  77 59 

CHAPTER  III 

Gr.\phics  of  the  /ji'-Plane 

section  x 

Plotting  Folylropics 

Plotting  the  Polytropic  Curve 61 

Exercises  78  to  S2 62 

Construction  of  the  Isothermal  Curve 64 

Exercise  8j 64 

SECTION   XI 

Graphical  Representation  of  ATF,  A/v,  and  AQ 

Graphical  Representation  of  lOxternal  Work 64 

Exercise  84 65 

Graphical  Representation  of  Internal  Energy 65 

Exercise  8§ 65 

Graphical  Representation  of  the  Change  in  Internal  Entrgj^ 66 

Exercise  86 66 

Exercise  87 68 

Graphical  Representation  of  the  Heat  Supplied 68 

Exercises  88  and  Sg 69 

CHAPTER  IV 

Compressors 

section  xii 

Single-Stage  Compressors 

Piston  Compressors 71 

The  Work  Required  to  Compress  and  Deliver 72 


CONTENTS  IX 

PAGE 

Exercise  go 74 

Exercises  gi  to  gj 75 

The  Effect  of  Clearance 75 

Exercises  g4  to  gS 76 

SECTION   XIII 

Compound  Compression 

Two-stage  Compression 78 

Exercises  gg  to  103 80 

Three-stage  Compression 80 

Exercise  104 81 

Turbo-compressor? 81 

Exercise  to^ 82 

CHAPTER  V 

Gas  Cycles 

section  xiv 

Introduction 

Definitions  and  General  Discussion 83 

section  XV 
Cycles  of  Hot-air  Engines 

The  Carnot  Cycle 86 

Exercises  106  and  107 89 

Exercises  inS  to  no go 

The  Stirling  Cycle 91 

Exercises  in  to  11  j 92 

The  Ericsson  Cycle 93 

Exercises  114  and  115 94 

The  Joule  Cycle 94 

Exercises  116  and  iiy 98 

Exercise  118 09 

Condition  Necessary  for  Greatest  Efficiency 100 

Cycles  Composed  of  Two  Pairs  of  Polytropics 10c 


X  CONTENTS 

SECTION  XVI 
Heal  Pumps  or  Rcjrigcraling  Machines 

PAGE 

Heat  Pumps 102 

Exercise  iig 102 

Exercise  120 104 

Thermodynamic  Coefficient  of  Performance 105 

Exercise  121 106 

The  Warming  Engine 106 

Exercise  122 108 

SECTION   XVII 

Cycles  of  Internal-combustion  Engines 

The  Brayton  Cycle log 

Exercise  123 109 

The  Otto  Cycle 109 

Analysis  of  the  Otto  Cycle no 

Exercises  124  to  I2y 112 

The  Diesel  Cycle 112 

Exercises  128  to  iji 114 

CHAPTER  VI 

The  Second  Law  of  Thermodynamics 
section  xviii 

The  First  Law 

Conservation  of  Energ>' 115 

Transformation  of  Energy 115 

The  First  Law  of  Thermodynamics 116 

SECTION   XIX 

The  Second  Law  of  Thermodynamics 

The  Second  Law 116 

Reversible  Processes  and  Cycles 117 

Exercise  ij2 119 


CONTENTS  XI 

PAGE 

Irreversible  Processes  and  Cycles up 

Carnot's  Principle 1 20 

Exercise  133 122 

The  Consequences  of  the  Second  Law  of  Thermodynamics 122 

SECTION   XX 

Kelvin's  Absolute  Sealc  of  Temperature 

Kelvin's  Absolute  Scale 123 

The  Efficiency  of  All  Reversible  Cycles 127 

SECTION   XXI 

The  Availability  of  Heat  Energy 

The  Availability  of  Heat  Energy 1 28 

The  Degradation  of  Heat  Energy 1 28 

Exercises  134  to  136 1 29 

Three  Types  of  "  Perpetual  Motion  " 1 29 


CHAPTER  Vn 

Entropy 

section  xxii 

Introduction 

The  Change  in  Entropy 131 

The  Entropy  of  a  Body  May  Be  Used  as  a  Coordinate 135 

section    XXIII 

Changes  in  Entropy  of  Ideal  Gases  during  Reversible  Processes 

Changes  in  Entropy  during  Reversible  and  Irreversible  Processes.  .  137 

Changes  in  Entropy  of  an  Ideal  Gas  during  Reversible  Processes.  .  138 

Exercises  137  to  i3g 139 

The  Temperature-Entropy  Diagram 140 

Reversible  Isopiestic  Processes  on  the  7"5-plane 140 

Exercises  140  to  142 143 


XU  CONTENTS 

PAGE 

Reversible  Isometric  Processes  on  the  r.s--planc 143 

Exercises  143  and  144 143 

Reversible  Isothermal,  Adiabatic,  and  Polytropic  Processes 144 

Exercise  145 144 

The  Temperature-Entropy  Plane 144 

Exercises  146  to  14Q 144 

The  Logarithm  Temperature-Entropy  Diagram 145 

Exercise  150 147 

SECTION   XXIV 

Gas  Cycles  on  the  Ts-plane 

Gas  Cj'cles  on  the  T^-pIane 147 

Exercises  151  to  i§4 148 

VAPORS 

CHAPTER  VIII 

Introduction 

section  xxv 

Properties  of  Vapors 

Properties  of  Water  and  Its  Vapor 149 

Exercise  155 151 

Exercise  156 152 

section  XXVI 

The  Characteristic  Surface  of  Vapors 

The  Characteristic  Surface 153 

Exercise  757 156 

SECTION  XXMI 

Isothermals 

Isothcrmals 156 

Critical  Pressure,  Volume,  and  Temperature 157 

Exercise  158 158 


CONTENTS  XUl 

PAGE 

The  Equation  of  Van  der  Waals 15^ 

Exercise  159 1 59 

CHAPTER  IX 

Formation  of  Vapors  at  Constant  Pressure 
section  xxviii 

Dry  Saluraled  Vapors 

Heat  Required  to  Warm  the  I-iquid 160 

Exercise  160 162 

The  Total  Heat  of  Dry  Saturated  Vapor 162 

The  Latent  Heat  of  Vaporization 162 

External  and  Internal  Latent  Heat 163 

The  Internal  Energj' 163 

Exercise  161 164 

Exercise  162 165 

The  Heat  Content  of  Dry  Saturated  Vapor 165 

The  Heat  Content  of  the  Liquid i<i6 

Exercise  163 ,  , 167 

SECTION  XXIX 

Wet  Vapors 

The  Quality  of  Wet  Vapors 167 

Exercise  164 168 

The  Total  Heat,  Internal  Energy,  and  Heat  Content  of  Wet  Vapor.  168 

Lines  of  Constant  Quality 168 

Exercise  165 169 

SECTION   XXX 

Superheated  Vapors 

Heat  Required  to  Superheat  Vapors  at  Constant  Pressure 169 

The  Internal  Energy  and  the  Heat  Content 170 

Exercises  166  atid  167 171 

Resumfi 171 


XIV  CONTENTS 

CHAPTER  X 

Entropy  of  Vapors 

section  xxxi 

Compulation  of  I  he  Change  in  Entropy 

PAGE 

The  Entropy  of  the  Liquid 174 

Exercise  168 175 

The  Entropy  of  Evaporation 175 

Exercise  i6g 175 

The  Entropjr  of  Dry  Saturated  Vapor 175 

Exercise  lyo 175 

The  Entropy  of  Wet  Vapor 176 

Exercise  171 176 

The  Entropy  of  Superheated  Vapor 176 

Exercise  172 176 

SECTION    XXXII 

The  Temperature-Enlro py  Diagrams  for   Vapors 

The  Temperature-Entropy  Diagram 176 

Lines  of  Constant  QuaHty 1 78 

Lines  of  Constant  Heat  Content 178 

Exercise  173 1 79 

Lines  of  Constant  Volume 179 

Exercise  174 179 

The  MolHer  Diagram 182 

Exercises  175  to  177 183 

Other  Diagrams 183 

CHAPTER  XI 

Changes  of  State  of  Vapors 
section  xxxiii 

Wet   Vapors 

Introduction 184 

Pressure  and  Temperature  Constant 185 

Exercises  178  and  179 187 


CONTENTS  XV 

PAGE 

Volume  Constant 187 

Exe.xises  180  to  184 189 

Entropy  Constant,  Reversible  Adiabatic  Change 189 

Exercise  18 j 190 

Exercises  J  86  to  i8g 191 

Exercise  igo 192 

Quality  Constant 192 

Exercise  igi 194 

SECTION   XXXIV 

Superheated  Vapors 

Pressure  Constant 195 

Exercises  ig2  and  igj 196 

Temperature  Constant 196 

Exercises  ig4  and  igs 197 

Volume  Constant 197 

Exercise  ig6 198 

Entropy  Constant,  Reversible  Adiabatic  Change iq8 

Exercise  igy 198 

Exercises  igS  to  200 199 

CHAPTER  XII 

Vapor  Cycles 
section  xxxv 

The  Carnal  Cycle 

The  Carnot  Cycle 2cx) 

Exercises  201  to  20 j 202 

The  Efficiency 202 

Exercise  204 203 

SECTION   XXXVT 

The  Rankine  Cycle 

The  Rankine  Cycle 203 

Exercises  205  to  2og 206 

The  Rankine  Cycle  with  Incomplete  Expansion 207 

Exercise  210 208 


XVi  CONTENTS 

SECTION   XXXVII 

Engine  Efficirncics 

PAGE 

Thermal  EfEciencies 209 

Exercise  211 211 

Exercises  212  to  214 212 

CHAPTER  XIII 

Flow  of  Fluids 
section  xxxviii 

Fundamental  Equations 

Definitions 213 

The  Equation  of  Continuity 214 

The  Equation  of  Energy 215 

Another  Form  of  the  Energy  Equation 217 

SECTION  xxx:x 

Adiahatic,  Friclionlcss  Flow 

The  Fundamental  Equations 218 

Exercises  215  to  217 220 

SECTION   XL 

Flow  thru  an  Orifice 

Introduction 221 

Exercises  21S  and  21  g 222 

The  Critical  Pressure 223 

Exercises  220  and  221 223 

Exercises  222  to  22s 225 

The  Discharge  thru  an  Orifice 225 

Fliegner's  Formula  for  Air 225 

Grashof's  Formula  for  Wet  Steam 227 

Exercise  226 228 

Napier's  Formula  for  Dry  Saturated  Steam 228 

The  Discharge  of  Vapors 229 

Exercise  22/ 229 


CONTENTS  XVU 

SECTION    XLI 

Flo'd'  thru  a  Nozzle  Including  the.  Effect  of  Friction 

PAGE 

The  Form  of  a  Nozzle 23 1 

Design  of  a  Nozzle,  Neglecting  Friction 236 

Exercise  228 237 

The  r^-plane "  237 

Exercises  22g  to  2J2 238 

Adiabatic  Flow  with  Friction 239 

Exercises  2jj  and  2^4 242 

Design  of  Nozzles  Incluc'ing  Friction 243 

Exercise  235 244 

SECTION   XLII 

Throttling 

Throttling 244 

Exercises  236  to  240 246 

Von  Linde's  Process  for  the  Liquefaction  of  Gases.  ^ 246 

The  Throttling  of  Wet  Vapors 247 

Exercises  241  to  244 247 

The  Loss  of  Availability  Due  to  Throttling 248 

Exercises  245  and  246 251 

Exercise  24J 253 

SECTION  XLIII 

Venturi  Meiers 

Venturi  Meters 253 

Exercise  24S 255 

Exercise  249 256 

SECTION   XLIV 

Floiv  thru  Pipes 

The  Hydraulic  Formula 256 

D'Arcy's  Formula 258 

Exercises  250  to  254 258 

A  More  Accurate  Formula 259 

Exercises  255  to  238 262 


XVlll  CONTENTS 

CHAPTER  XIV 

General  IiIquation  of  Thermodynamics 

section  xlv 

Differential  Expressions  for  the  Heat  Supplied 

PAGE 

The-Heat  Absorbed  during  any  Change  of  State 264 

Relations  between  the  Thermal  Capacities 267 

Exercise  2^g •  268 

Exercises  260  and  261 270 

SECTION   XLVI 

Exact  Differentials 

Definitions 271 

Exercises  262  and  263 273 

The  Test  for  an  Exact  Differential 274 

Exercise  264 275 

SECTION  XLVII 

The  Differential  Equations  of  Thermodynamics 

Entropy  as  a  Coordinate 275 

Thermodynamic  Potentials 278 

Exercise  265 280 

Maxwell's  Relations 280 

Exercises  266  and  26y 281 

Exercise  268 282 

A  General  Expression  for  Cp—Cv 282 

Exercises  26g  and  270 283 

The  Heat  Supplied 283 

Exercise  271 283 

Exercise  272 284 

Clapeyron's  Formula , ; 284 

Exercise  273 285 

Other  Important  Relations 286 

Exercises  274  to  276 286 

Problems  for  Review 289 

Exercises  277  to  382 289 

Answers  to  All  Exercises 305 

Index 311 


THERMODYNAMICS 


INTRODUCTION 

Thermodynamics  is  the  science  which  deals  with  the 
relation  between  heat  and  other  forms  of  energy.  As  most 
physical,  chemical,  and  biological  processes  involve  energy 
transformations  involving  heat,  thermodynamics  is  a  most 
fundamental  science. 

In  this  text-book  the  subject  matter  will  be  limited  to 
the  relation  of  heat  to  mechanical  energy  in  both  its  po- 
tential and  kinetic  forms.  It  will  include  the  mechanics 
of  gases  and  vapors  with  special  emphasis  on  technical 
or  engineering  applications. 

Energy. — A  body  possesses  energy  when  it  can  do  work, 
when  it  can  move  itself  or  other  bodies  against  resistances. 
The  various  forms  in  which  energy  is  known  to  exist  are 

1.  Kinetic  energy 

2.  Potential  energy 

3.  Heat  energy 

4.  Radiant  energy 

5.  Electric  energy 

6.  Magnetic  energy' 

7.  Chemical  energy. 

Heat  is  a  Form  of  Energy. — It  has  been  experimentally 
demonstrated   that   other    forms   of   energy   may   be   con- 


2  INTRODUCTION 

verted  into  heat  and  that  heat  may  be  transformed  into 
other  forms  of  energy.  Thus  heat  must  be  a  form  of 
energy. 

It  is  useful  to  conceive  a  way  in  which  this  heat  energy 
may  reside  in  a  body.  As  a  working  hypothesis  it  is  usual 
to  postulate  the  molecular  constitution  of  matter  and  to 
point  to  the  molecules  as  a  storage  place  of  heat  energy. 
The  Greek  philosophers  originated  the  idea  that  matter 
consists  of  ultimate  particles  or  molecules  separated  by 
interspaces.  By  means  of  this  hypothesis  many  physical 
properties  of  matter  such  as  compressibility,  solubility, 
etc.  are  readily  explained;  moreo^'er,  the  laws  of  chemical 
combination  are  thereby  explained  with  wonderful  sim- 
plicity. 

Internal  Kinetic  Energy. — The  molecules  of  which 
matter  is  assumed  to  be  composed  are  conceived  to  be  in 
constant  vibratory  motion.  Their  mean  velocity  deter- 
mines their  kinetic  energy  and  this  mean  velocity  is 
assumed  to  increase  with  increasing  temperature.  The 
molecules  of  a  body  whose  temperature  is  high  may  be 
conceived  as  producing  the  waves  in  the  ether  (which 
constitute  radiant  energy)  owing  to  their  rapid  vibratory 
motion. 

The  higher  the  temperature  of  a  body  the  greater  will 
be  its  store  of  internal  kinetic  energy. 

Internal  Potential  Energy. — The  molecules  of  which 
matter  is  composed  are  conceived  to  exert  an  attraction 
for  each  other.  This  mutual  attraction  ex-plains  the  co- 
hesion which  matter  exhibits.  If  the  mean  distance  between 
molecules  is  increased  then  work  must  be  done  in  over- 
coming this  attraction.     Energy,  equal  in  amount  to  the 


INTRODUCTION  •         3 

work  done,  is  stored  within  the  body  as  internal  potential 
energy. 

A  change  in  the  mean  distance  between  the  molecules 
may  manifest  itself  as  a  change  in  volume  or  as  a  change 
in  ph)'sical  state.  In  either  case  a  rearrangement  of  the 
molecules  occurs. 

External  Work. — The  heat  supplied  to  a  body  from  an 
external  source  does  not  reside  wholly  in  the  molecules 
of  the  body  as  internal  kinetic  and  potential  energies. 
Expansion  usually  accompanies  the  absorption  of  heat. 
Energy  is  necessary  to  do  the  work  required  in  overcoming 
any  external  pressure  exerted  upon  the  body.  This  energy 
must  be  supplied  to  the  body  in  the  form  of  heat  energy. 

To  compute  the  external  work  done  in  overcoming  an 
external  pressure,  p,  acting  upon  the  expanding  body  note 
that  the  force  on  a  differential  element,  dA,  of  the  surface 
of  the  body  is  pdA.  If  this  force  is  displaced  normally  to 
the  surface  thru  a  distance  ds  the  work  done  is  pdAds. 
As  dAds  represents  a  differential  volume  of  the  increase 
in  size  of  the  body  which  may  be  denoted  hy  dV  we  have 

as  the  external  work  performed,    I  pdV. 

The  Conservation  of  Energy .^The  heat  absorbed  by 
a  body  from  some  external  source  must  be  wholly  accounted 
for,  as  energy  has  never  been  known  to  be  either  created 
or  destroyed. 

In  order  to  establish  an  equation  expressing  the  con- 
servation of  energy  as  applied  to  the  above  described  forms 
of  energy  let 

AQ  be  the  heat  absorbed  by  a  body  from  external  sources, 
AP   be  that  part  of  A()  which  is  stored  in  the  body  as  an 
increase  in  internal  potential  energy, 


4  INTRODUCTION 

Ai<C  be  that  part  of  AQ  which  is  stored  in  the  body  as  an 
increase  in  internal  kinetic  energy, 

AW  be  that  part  of  AQ  which  must  be  expended  in  over- 
coming the  external  pressure  acting  upon  the  body 
and  which  is  stored  as  external  potential  energy, 

then   for  our   purpose  the  fundamental  equation  expressing 

the  conservation  of  energy  is 

AQ  =  AP+AK+AW. 

To  illustrate  the  meaning  of  these  symbols  consider 
the  heating  of  water.  Here  the  increase  in  volume  is  very 
slight  and  therefore  AW  is  very  small.  As  no  change 
in  physical  state  occurs  and  as  the  mean  distance  between 
the  molecules  is  not  otherwise  greatly  increased  by  expan- 
sion AP  is  also  small.  The  temperature  does  increase  so 
that  AK  is  relatively  large  and  practically  equals  the  whole 
of  AQ.  Thus  the  heat  supplied  is  practically  stored  as 
increased  internal  kinetic  energy  due  to  an  increase  in 
the  mean  vibratory  velocity  of  the  molecules. 

Consider  now  the  conversion  of  boiling  water  into  steam. 
Here  the  temperature  remains  constant  so  that  AK  is 
zero.  AW  and  AP  however  are  very  large  owing  to  the 
large  increase  in  volume  and  to  the  change  in  physical 
state  which  calls  for  a  rearrangement  of  the  molecules, 
the  mean  distance  between  molecules  having  greatly  in- 
creased. 

The  Units  to  be  Used. — The  units  used  by  English- 
speaking  engineers  for  measuring  the  energies  above  dis- 
cussed are  the  foot-pound  for  the  unit  of  work  and  energy 
and  the  British  thermal  unit  for  the  unit  of  heat. 


INTRODUCTION  5 

The  British  thermal  unit  (B.t.u.)  is  defined  as  y g-^  of  the 
heat  which  is  absorbed  by  a  mass  of  one  pound  of  water 
while  its  temperature  increases  from  32  to  212°  F,  the 
pressure  remaining  constant  and  equal  to  the  standard 
atmosphere.  This  unit  is  sometimes  called  the  mean 
B.t.u.  and  it  equals  the  heat  required  to  change  the  tem- 
perature of  one  pound  of  water  from  63  to  64°  F  under 
atmospheric  pressure. 

The  First  Law  of  Thermodynamics. — The  first  law  of 
thermodynamics  states  that  not  only  can  heat  be  converted 
into  other  forms  of  energy  (and  vice  versa)  but  that  the 
complete  transformation  of  a  definite  quantity  of  heat  will 
always  yield  a  definite  quantity  of  mechanical  energy. 

Numerous  experiments  have  led  to  the  conclusion  that 
777.6  foot-pounds  of  mechanical  energy  always  yield  one 
B.t.u.  of  heat. 

This  constant,  which'  for  purposes  of  engineering  cal- 
culations may  be  assumed  to  be  778,  will  for  convenience 
be  represented  by  the  letter  /  in  honor  of  Joule  who  in 
1843  made  the  first  experiments  leading  to  a  determination 
of  this  constant. 

Thermal  Capacity.— The  heat  required  to  raise  the  tem- 
perature of  a  body  one  degree  F  is  called  the  thermal 
capacity  of  the  body. 

The  thermal  capacity  of  a  body  may  vary  with  each 
degree  of  the  thermometric  scale  and  even  for  the  same 
range  of  temperature  the  thermal  capacity  of  a  given 
body  need  not  be  constant  under  all  conditions.  It  should 
be  remembered  that  the  heat  supplied  may  be  stored  in 
three  ways:  (i)  as  internal  kinetic  energy  manifested  to 
our  senses  as  a  change  in  temperature,  (2)  as  internal  po- 


6  INTRODUCTION 

tential  energy  manifested  as  a  change  in  volume  or  as  a 
change  in  physical  state,  (3)  as  external  work  performed 
in  overcoming  external  pressure.  At  least  the  third  manner 
in  which  heat  may  be  absorbed  by  a  body  must  yield  vary- 
ing results  with  varying  external  conditions.  These  con- 
ditions have  no  connection  with  the  nature  of  the  body. 

If  AQ  represents  the  heat  absorbed  under  certain  con- 
ditions with  an  increase  in  the  temperature  of  the  body 
from  h  to  t2°  F,  then  the  mean  thermal  capacity  of  tht 

body  equals  — ~  for  this  range  of  temperature  and  under 
^2 — h 

the  existing  conditions. 

The  instantaneous  value  of  the  thermal  capacity  of  a 

body  at  a  temperature  t°  F  would  be  represented  by 


dQ 

dt' 


If  the  thermal  capacity  be  divided  by  the  mass  of  the 
body  in  pounds  the  thermal  capacity  per  unit  of  mass  is 
obtained. 

For  water  at  63.5°  F  and  under  standard  atmospheric 
pressure,  the  thermal  capacity  per  unit  mass  is  by  defini- 
tion one  B.t.u. 

Specific  Heat. — The  specific  heat  of  a  substance  at  a 
given  temperature  and  under  given  external  conditions  is 
the  ratio  of  the  thermal  capacity  per  unit  mass  of  this 
substance  at  the  given  temperature  and  under  the  given 
conditions  to  the  thermal  capacity  per  unit  mass  of  water 
under  some  chosen  standard  temperature  and  external 
conditions. 


INTRODUCTION  7 

If  the  chosen  temperature  and  pressure  for  water  are 
63.5°  F  and  atmospheric  pressure  then  its  thermal  capacity 
per  unit  mass  is  unity  and  the  specific  heat  of  the  substance 
becomes  numerically  equal  to  its  thermal  capacity  per 
unit  mass  under  the  given  conditions. 

The  heat  absorbed  by  a  given  quantity  of  a  substance 
would  thus  be  computed  from  the  continued  product  of  its 
specific  heat  under  the  existing  conditions,  its  mass,  and 
the  change  in  temperature. 


GASES 

CHAPTER  I 
THE  LAWS  OF  IDEAL  GASES 

Section  I 
JOULE'S,   BOYLE'S,   AND    CHARLES'   LAW 

The  laws  about  to  be  discussed  are  obeyed  more  or  less 
exactly  by  all  gases.  The  more  the  conditions  under  which 
the  gases  exist  are  removed  from  the  conditions  approaching 
liquefaction  the  more  closely  do  the  gases  obey  these  laws. 
Thus  at  ordinary  temperatures  and  pressures  water  vapor 
does  not  obey  the  laws  even  approximately,  ammonia  and 
carbon  dioxide  may  as  a  first  approximation  be  assumed 
to  follow  the  laws  of  ideal  gases,  while  hydrogen,  oxygen, 
nitrogen,  and  air  obey  these  laws  with  sufficient  accuracy 
for  all  engineering  purposes. 

Any  hypothetical  gas  which  would  obey  absolutely  the 
following  laws  is  called  an  ideal  gas  (also  a  perfect  gas). 
Actual  gases  approach  this  ideal  condition  more  closely 
as  their  molecules  become  more  and  more  separated  and 
thus  exert  less  and  less  influence  upon  each  other.  An 
ideal  gas  must  thus  be  conceived  to  be  so  attenuated  that 
intermolecular  forces  no  longer  exist.  This  conception  leads 
to  the  first  law  of  ideal  gases. 

9 


lO  THERMODYNAMICS 

Joule's  Law. — In  an  ideal  gas  no  change  in  internal 
potential  energy  occurs  when  the  gas  absorbs  or  rejects 
heat.  Thus  in  an  ideal  gas  any  change  in  internal  energy 
must  be  a  change  in  internal  kinetic  energy  which  mani- 
fests itself  as  a  change  in  temperature. 

In  our  notation  we  have  for  ideal  gases 

AP  =  0. 

To  test  this  law  Joule  devised  the  following  experiment 
not  knowing  that  Gay-Lussac  had  already  performed  it 
in  a  slightly  different  way.  Joule  connected  two  large 
receivers  by  means  of  a  pipe  closed  with  a  valve.  In  one 
of  these  receivers  he  compressed  air  to  22  atmospheres; 
the  other  he  exhausted.  The  whole  apparatus  was  im- 
mersed in  a  well  insulated  water-bath  whose  temperature 
could  be  accurately  determined.  The  experiment  consisted 
in  noting  the  change  in  temperature  due  to  the  flow  of  air 
from  one  receiver  into  the  other  after  the  valve  was  opened. 

No  change  in  temperature  was  detected. 

In  the  notation  of  our  fundamental  energy  equation 

AQ  =  AP-\-AK-\-AW 

this  experiment  was  so  devised  that 

AQ  =  o,  for  no  heat  was  supplied  to  the  air  and 

APT =0,  for  no   external  work  was  done  during   the  flow 

of    the   air   from   one    receiver   into   the   other. 

Moreover   the    experiment    showed   that   under 

the  existing  conditions 
AK  =  o,  for    no    change    in   temperature  occurred.      Thus 
AP  must  be  equal  to  zero. 


THE    LAWS    OF    IDEAL    GASES  II 

If  the  receivers  in  Joule's  experiment  are  each  immersed 
in  a  separate  water  bath,  then  considerable  cooling  would 
be  noticed  in  the  bath  surrounding  the  compressed  air 
receiver  and  the  temperature  of  the  bath  surrounding  the 
originally  exhausted  receiver  would  be  correspondingly 
increased.  This  is  explained  by  the  fact  that  the  air  leaving 
the  high  pressure  receiver  does  so  with  considerable  velocity 
and  therefore  kinetic  energy.  This  kinetic .  energy  can  be 
derived  only  from  energy  already  in  the  system.  Thus 
heat  energy  equal  in  amount  to  the  kinetic  energy  dis- 
appears with  a  corresponding  drop  in  temperature.  Simi- 
larly the  kinetic  energy  carried  over  into  the  other  receiver 
is  transformed  by  impact  and  friction  into  heat  energy 
with  a  corresponding  rise  in  temperature. 

Boyle's  Law. — ^At  constant  temperature  the  volumes  of 
a  given  mass  of  an  ideal  gas  are  inversely  proportional 
to  che  corresponding  pressures. 

Thus  ir^:r' 

V2    pi 

or  p\  Vi  =  p2  V2  =  a  constant, 

when  /  is  constant. 

Law  of  Charles. — At  constant  pressure,  each  change  of 
one  degree  Fahrenheit  in  temperature  causes  a  change  in 
the  volume  of  an  ideal  gas  at  32°  F  equal  to  ^l^  of  the 
volume  at  32°  F,  and 

At  constant  volume,  each  change  of  one  degree  Fahrenheit 
causes  a  change  in  the  pressure  at  32°  F  equal  to  ^J  ,^  of  the 
pressure  at  ^2°  F. 

To  express  the  two  forms  of  this  law  in  symbols,  let 
jjj  be  represented  by  a  for  convenience,  and  let  the  zero 


12  THERMODYNAMICS 

subscript  refer  to  the  conditions  existing  at  32°  F,  so  that 
po  and  Vo  represent  the  pressure  and  the  volume  of  the  mass 
of  ideal  gas  when  the  temperature  is  32°  F.  Then  the 
law  of  Charles  may  be  written 

V=Vo+Voa(t-s2), 

when  p  is  constant  and 

p  =  po-\-poa(t-7,2), 

when  V  is  constant,  where  p  and  V  are  the  pressure  and 
the  volume  when  the  temperature  is  /°  F. 

Absolute  Zero. — The  above  symbolic  expressions  of  the 
law  of  Charles  may  be  simplified  by  using,  instead  of  the 
Fahrenheit  scale  of  temperature,  a  scale  having  the  same 
degrees  but  whose  zero  is  at  the  lowest  conceivable  tem- 
perature, the  absolute  zero. 

The  kinetic  theory  of  gases  shows  that  the  pressure 
which  a  gas  exerts  may  be  explained  as  the  result  of  the 
bombardment  of  the  walls  of  the  enclosing  vessel  by  the 
molecules  of  the  gas.  The  impact  and  therefore  the  pressure 
exerted  depends  upon  the  velocity  of  the  impinging  mole- 
cules. This  velocity  in  turn  depends  upon  the  temperature 
of  the  gas.  Thus  no  temperature  means  no  vibratory 
velocity  and  no  pressure.  To  find  the  absolute  zero  of 
temperature  on  the  Fahrenheit  scale  we    may  thus   place 

the  pressure  p  in 

p  =  po+poa(t—S2) 

equal  to  zero  and  solve  for  i. 

We  find  /=— 460°  F.  Therefore  the  absolute  zero  lies 
492  Fahrenheit  degrees  below  the  freezing  point  of  water. 


THE    LAWS    OF   IDEAL   GASES 


13 


Temperature  referred  to  this  absolute  zero  is  called  ab- 
solute temperature  and  will  always  be  denoted  by  capital  T. 

Thus  r=46o+/ 

where  /  is  the  temperature  on  the  Fahrenheit  scale. 

Placing  /=r— 460  in  the  equations  expressing  the  law 
of  Charles  we  obtain 

F=  VoaT 

and  p  =  poaT. 

The  General  Law  for  Ideal  Gases. — A  single  law  in- 
cluding Boyle's  law  and  both  forms  of  Charles'  law  is 
desirable. 

To  recapitulate  we  have 

at  constant  temperature 

pV  =  poVo  =  a.  constant, 
at  constant  pressure 

—  =a\o  =  a  constant, 

at  constant  volume 

P 

—  =  apo    =a  constant. 

A  single  equation  combining  these  three  equations  may 
be  obtained  by  equating  the  product  of  their  left-hand 
members  to  the  product  of  their  right-hand  members, 
as  follows: 

^=a-Po'V.\ 


14  THERMODYNAMICS 

As  a= —  and  as  T  =  A6o-\-t,  whence  To =402,  we  have 
492 

pV^poVo 

T       To  ' 

Another  method  of  obtaining  this  general  relation  between 
the  pressure,  the  volume,  and  the  temperature  of  any 
given  mass  of  gas  under  any  conditions  and  the  corre- 
sponding values  at  32°  F  consists  in  assuming  an  intermediate 
step  in  the  transition  as  illustrated  in  Fig.  i.     This  inter- 


FlG.   I. 

mediate  step  is  necessary,  because  the  application  of  either 
Boyle's  or  Charles'  law  requires  in  each  case  a  constant 
condition.  Thus  we  may  maintain  T  constant  during  the 
change  from  A  to  B  (Fig.  i)  during  which  p  becomes  po 
and  V  changes  to  V.  During  the  second  stage  of  the 
transition  po  must  now  be  maintained  constant,  while 
V  changes  to  Vq  and  T  to  To- 

Applying  Boyle's  law  during  the  change  from  A  to  B, 
we  have 

pV=PoV\ 


THE    LAWS    OF    IDEAL   GASES  1 5 

From  Charles'  law  applied  to  the  change  from  B  to 
C  we  have 

T     To 

By  the  elimination  of  V  from  these  equations 

pV^PoVo 
T        To  ' 

Exercise  i.  Deduce  the  last  equation  assuming  the  pressure 
to  remain  constant  during  the  first  step  of  the  transition  and 
the  temperature  to  remain  constant  during  the  second  part  of 
the  transition. 

In  the  above  equations  V  represents  the  volume  of  a 
given  mass  of  m  pounds  of  gas.  Let  v  represent  the  volume 
of  one  pound  of  the  gas,  then 

V  =  7nv,  and  of  course  \'o=>f''Vo, 

and  we  may  write 

pvm  _  poVo 
T    ~  To  ""' 

or  ^  =  mR 

where  i?  is  a  constant  for  any  given  gas  and 

R  is  called  the  gas  constant.  Its  value  depends  not 
only  upon  the  nature  of  the  gas  considered  but  also  upon 
the  units  selected  in  the  measurement  of  p,  V,  T,  and  m. 


1 6  THERMODYNAMICS 

English-speaking  engineers  always  measure  p,  the  abso- 
lute pressure,  in  pounds  per  square  foot,  V ,  the  volume,  in 
cubic  feet,  T,  the  absolute  temperature,  in  Fahrenheit 
degrees,  m,  the  mass  of  the  gas,  in  pounds. 

Exercise  2.  The  weight  of  atmospheric  air  at  normal  tem- 
perature and  pressure  is  0.08071  pound  per  cubic  foot.  Find 
the  gas  constant  for  air. 

Exercise  3.  At  what  temperature  will  50  pounds  of  air  occupy 
60  cubic  feet  when  under  an  absolute  pressure  of  205  pounds 
per  square  inch?     {R  for  air  53.3.) 

Exercise  4.  A  cylindrical  tank  (capacity  3  cubic  feet)  con- 
tains 20  pounds  of  air  at  60°  F.  To  what  internal  fluid  pres- 
sure is  the  tank  subjected? 

Section  II 
THE   GAS   CONSTANT   OF   ANY   IDEAL    GAS 

In  order  to  determine  the  gas  constant  for  any  ideal 
gas  in  terms  of  the  molecular  weight  of  the  gas,  Avogadro's 
law  is  used. 

Avogadro's  Law  states  that  at  the  same  temperature 
and  pressure  equal  volumes  of  all  ideal  gases  contain  the 
same  number  of  molecules. 

Thus  the  weights  of  equal  volumes  of  two  ideal  gases 
must  be  proportional  to  their  molecular  weights.  As  the 
weight  of  oxygen  (molecular  weight  32)  is  found  to  be 
0.08922  pound  per  cubic  foot  at  n.t.p.  the  specific  weight 
of  any  other  ideal  gas  whose  molecular  weight  is  m  is 


(0.08922)!  —  I  at  n.t.p. 


THE    LAWS    or    IDEAL    GASES  1 7 

The  volume  of  one  pound  of  any  gas  is  thus  -. -r-; 

cubic  feet  when  under  an  absolute  pressure  of  14.7  pounds 
per  square  inch  and  at  a  temperature  of  32°  F. 
The  gas  constant  of  any  gas  is  therefore 

To     492 X 0.0892 2 Xm       m    ' 

where  n  is  the  molecular  weight  of  the  gas. 

Exercise  5.  Show  that  the  weight  of  a  cubic  foot  of  any 
gas  at  n.t.p.  in  pounds  is 

3S8* 

Exercise  6.  Compute  R  for  nitrogen. 

Exercise  7.  A  tank  (capacity  5  cubic  feet)  contains  2  pounds 
of  acetylene.  At  what  temperature  will  the  pressure  in  the 
tank  reach  100  pounds  per  square  inch  gage? 

Exercise  8.  Two  receivers,  capacities  200  and  100  cubic  feet, 
are  maintained  at  temperatures  of  150°  F  and  80°  F  respect- 
ively. These  receivers  are  connected  by  a  pipe  and  contain 
1 20  pounds  of  air. 

(a)  Find  the  pressure  in  the  system. 

(b)  What  weight  of  air  does  the  larger  receiver  contain? 
Exercise  9.  Two  tanks,  each  having  a  capacity  of  3  cubic 

feet,  are  filled  at  1500  pounds  per  square  inch  gage,  one  with 
oxygen,  the  other  with  hydrogen,  both  at  70°  F.  If  these 
tanks  are  of  equal  weight  when  empty,  find  the  difference  in 
their  charged  weights. 


jg  THERMODYNAMICS 

Section  III 
THE    GAS   CONSTANT   OF    GASEOUS    MIXTURES 

To  compute  the  gas  constant  of  a  mixture  of  ideal  gases 
Dalton's  law  is  used. 

Dalton's  law  states  that  the  pressures  exerted  by  a 
mixture  of  ideal  gases  upon  the  walls  of  the  containing 
vessel  equals  the  sum  of  the  pressures  due  to  each  of  the 
constituent  gases,  provided  they  each  in  turn  alone  occupied 
the  containing  vessel  at  the  temperature  of  the  mixture. 

The  law,  pV=mRT,  is  apphcable  to  the  mixture  as 
well  as  to  each  individual  gas.  It  remains  to  find  the 
value  of  R  for  the  mixture  in  terms  of  the  i?'s  of  the  indi- 
vidual gases  composing  the  mixture. 

Let  the  mass  w  of  the  mixture  be  composed  of  the  masses 
mi,  W2,  .  .  .  of  the  constituent  gases  and  let  the  pressure 
of  the  mixture  be  p  and  equal  to  pi-\-p2-\--  •  .,  the  sum 
of  the  pressures  which  each  gas  occupying  the  whole  vol- 
ume of  the  mixture  would  individually  exert,  this  in  accord- 
ance with  Dalton's  law. 

Then  if  Ri,  R2,  .  .  .  are  the  gas  constants,  of  the  con- 
stituent gases  we  have  from  the  general  equation  of 
gases  under  the  assumption  that  each  gas  alone  occupies 
the  original  space  of  the  mixture  at  the  temperature  of 
the  mixture, 

pi]'  =  niiRiT 

p2V  —  m2R2T,  etc. 

Adding  these  equations  we  obtain 

pV={miRi+7n2R2-\-  .  .  .)T. 


THE    LAWS    OF    IDEAL    GASES  I9 

Comparing  this  equation  with  the  equation 

pV=mRT 

applied  to  the  mixture  and  in  which  R  is  the  gas  constant 
of  the  mixture  we  see  that 

mR  =  miR]_-\-m2R2-\--  -  • 

m  m 

Here  —  ,  — ■,  •  .  ■  represent  the  parts  by  weight  of  the 
w     m 

constituent  gases  forming  the  mixture. 

Exercise  io.  Show  that  the  pressure  due  to  any  one  gas 
of  a  mixture  may  be  represented  by 

mn  Rn  , 
^"=m'R^- 

Exercise  it.  Air  contains  23.6  per  cent  of  oxygen  apd 
76.4  per  cent  of  nitrogen  by  weight. 

(a)  Compute  the  gas  constant  for  air  by  means  of  the  gas 
constants  of  oxygen  and  of  nitrogen. 

(h)  What  part  of  the  30  inches  of  mercury  representing  nor- 
mal atmospheric  pressure  is  due  to  the  oxygen,  and  what  part 
is  due  to  the  nitrogen  of  the  atmosphere? 

Exercise  12.  Compute  the  apparent  molecular  weight  of 
air  by  means  of  the  gas  constant  computed  in  Exercise  11  {a). 

Exercise  13.  Compute  the  weight  of  a  cubic  foot  of  air  at 
n.t.p.  by  means  of  the  result  obtained  in  Exercise  12. 

Exercise  14.  Analysis  of  a  flue  gas  shows  12  parts  of  COj, 
6  parts  of  O2,  and  82  parts  of  N2  by  volume. 

Compute  {a)  the  gas  constant, 

{b)  the  apparent  molecular  weight, 

(r)  the  specific  weight  (pounds  jht  cubic  foot)  at  n.t.p.  of 
this  mixture. 


20  THERMODYNAMICS 

Section  IV 
THE    SPECIFIC   HEAT    OF   GASES 

It  follows  from  the  units  and  the  standard  conditions 
described  in  the  Introduction  that  the  specific  heat  of 
any  substance  is  numerically  equal  to  its  thermal  capacity 
per  unit  mass. 

If  c  represents  the  specific  heat  under  certain  conditions 
of  temperature  and  pressure  then  the  heat  AQ  absorbed 
by  a  mass  m  of  the  given  substance  may  be  computed  as 
follows 


Ae=/, 


mcdl. 

If  c  is  or  is  assumed  constant  then 
A(2  =  mcA^. 

The  heat  absorbed  by  a  body  may  have  other  effects 
besides  that  indicated  by  a  change  in  temperature.  It 
may  do  external  work.  The  heat  absorbed  computed  by 
means  of  a  specific  heat  must  be  interpreted  as  including 
not  only  the  heat  required  to  increase  the  temperature 
even  tho  it  is  computed  by  means  of  a  change  in  tempera- 
ture but  also  all  the  heat  absorbed  and  stored  or  utilized 
in  other  ways  as  in  doing  external  work. 

Thus  the  specific  heat  of  a  body  may  have  various  values 
even  for  the  same  range  of  temperature  depending  upon 
the  external  conditions. 

For  solids  and  liquids  the  specific  heats  will  vary  only 
slightly,  for  a  gas  the  specific  heats  may  have  any  value 
from  plus  to  minus  infinity  for  the  same  range  in  tem- 
perature, under  various  external  conditions. 


THE    LAWS    OF   IDEAL   GASES  21 

The  specific  heat  of  a  gas  depends  not  only  upon  the 
nature  of  the  gas  but  also  upon  the  conditions  under  which 
the  heat  is  supplied  to  the  gas.  These  conditions  deter- 
mine how  much  of  the  heat  energy  must  be  used  in  doing 
external  work  during  the  heating  of  the  gas. 

The  energy  stored  in  an  ideal  gas  must  be  stored  wholly 
as  internal  kinetic  energ}'  (Joule's  law),  so  that  even  with 
increasing  volume  and  the  accompanying  increase  in  the 
mean  distance  between  the  molecules  of  an  ideal  gas  no 
change  in  the  internal  potential  energy  occurs. 

The  energy  equation  for  an  ideal  gas  is  therefore 

AQ=AK+AW. 

The  Specific  Heat  at  Constant  Volume.— If  a  gas  is 
heated  at  constant  volume  no  external  work  can  be  per- 
formed during  the  absorption  of  the  heat. 

Therefore  AWv  =  o, 

and  AQv  =  AKv, 

where  the  subscripts  v  are  used  to  indicate  change  at  con- 
stant volume.  Under  these  conditions  the  whole  heat  energy 
absorbed  is  stored  in  the  gas. 

Let  Cv  represent  the  specific  heat  of  the  gas  at  constant 
volume  then 


^Qv=^nj^ 


CvdT. 

Ti 


It  can  be  shown  mathematically  (see  Exercise  275)  that 
the  Cv  for  any  gas  obeying  the  law  pV=mRT  must  have 
the  same  value  no  matter  at  what  pressure  or  at  what 
volume  the  gas  may  be  heated.     But  Cp  may  vary  with  the 


22  THERMODYNAMICS 

temperature  and  experiments  have  shown  that  the  specific 
heats  of  all  gases  are  linear  functions  of  the  temperature. 
Thus  the  changing  value  of  Cp  for  any  gas  may  be  rep- 
resented by 

Cv  =  a-\-bT, 

where  a  and  h  are  empirical  constants. 

Moreover  it  has  been  shown  experimentally  that  equal 
volumes  of  diatomic  gases  have  the  same  thermal  capacity 
at  the  same  temperature.  As  the  masses  of  equal  volumes 
of  diatomic  gases  are  proportional  to  the  molecular  weights 
of  these  gases  (Avogadro's  law)  it  follows  that  the  thermal 
capacities  of  masses  of  diatomic  gases  which  are  propor- 
tional to  their  molecular  weights  must  be  equal  or 

where  {cv)\,  {c^2  •  •  -  are  the  specific   heats   at   constant 

volume,  and  Ati,  M2,  •  •  •  are    the  corresponding  molecular 

weights  of  the  diatomic  gases. 

Langen  has  shown  that  for  diatomic  gases  such  as 
H2,  molecular  weight=    2.016 
N2,  molecular  weight  =28.08 
O2,  molecular  weight  =3 2. 00 
CO,  molecular  weight=  28.00 
NO,  molecular  weight  =  30. 04 

and  for  mixtures  of  these  gases, 

MCi,=4.625+o.ooo588r, 

where  T  is  expressed  in  absolute  Fahrenheit  degrees. 
For  carbon  dioxide  (jj.  =  44)  Langen  found 

^iC5=6.774-fo.oo2ior 


THE    LAWS    OF    IDEAL   GASES  23 

and  for  highly  superheated  steam  (/x=  1 8.01 6) 

/xc„  =  4.72+o.oo2387' 

where  T  is  in  absolute  Fahrenheit  degrees. 

These  experimental  results  show  that  the  more  a  gas 
approaches  the  condition  of  the  so-called  permanent  gases 
the  less  does  the  specific  heat  at  constant  volume  vary 
with  the  temperature. 

Ct,  for  any  ideal  gas  is  assumed  to  remain  constant 
for  all  temperatures. 

The  c„  of  diatomic  gases  may  be  considered  constant 
between  32  and  400°  F  for  engineering  computations,  and 
may  be  taken  as 

0.1 7 1  for  air 

2.42    for  H2 

0.174  for  N2 

0.155  fo''  O2 

0.172  for  CO. 

Exercise  15.  (a)  At  what  temperature  is  Cv  for  air  equal 
to  0.171? 

{b)  What  is  the  percentage  increase  in  the  Cv  of  any  diatomic 
gas  between  32  and  400°  F? 

Exercise  16.  If  fxCv  =  a-\-hT,  where  a  and  /;  are  constants, 
compute  the  heat  required  to  change  the  temperature  of  m 
pounds  of  the  gas  from  Ti  to  T2  degrees  at  constant  volume. 
What  is  the  mean  specific  heat  of  the  gas  between  Ti  and  Tz 
degrees? 

The  Specific  Heat  at  Constant  Pressure. — When  a  gas 
is  heated  under  constant  pressure  its  volume  increases 
according  to  the  law  pV  =  mRT  and  external  work  must  be 


24  THERMODYNAMICS 

performed  in  overcoming   the  constant  pressure   p.     This 
work  equals 

where    the    subscript    p    indicates    constant    pressure    and 
the  result  is  expressed  in  heat  units. 

During  this  change  of  volume  the  temperature  of  the  gas 
has  increased  from  Ti  to  T2°.  Thus  the  internal  energy 
of  the  gas  has  increased,  and  according  to  Joule's  law  its 
internal  kinetic  energy  only  has  increased.  As  this  increase 
in  internal  kinetic  energy  is  independent  of  the  change  in 
volume  and  depends  only  on  the  change  in  temperature 
we  see  that  Ai^^,  is  the  same  as  any  Ai^C  occurring  under 
any  other  external  conditions.     We  may  put 

CidT. 

Ti 

If  Cv  is  constant  as  it  is  for  an  ideal  gas  then 

AKp=A.Kv  =  niCviT2—Ti). 

The  total  heat  energy  added  to  an  ideal  gas  changing 
its  state  under  constant  pressure  is  thus 

AQp  =  AKp-{-AWp  =  mc,{T2-Ti)-\-jPiV2-Vi). 

The  heat  absorbed  by  an  ideal  gas  under  constant  pres- 
sure may  also  be  computed  by  means  of  its  specific  heat 
at  constant  pressure,  thus 

and  as  Cp  is  constant 

AQp=inCp{T2—Ti). 


THE    LAWS    OF    IDEAL   GASES  2$ 

Note  that  the  use  of  Cp  leads  directly  to  the  value  of 
AQp  computed  above  in  two  steps  by  means  of  the  energy 
equation. 

The  Relation  between  Cp  and  Cv. — As  the  AQp's  just 
computed  in  two  ways  must  be  equal  we  may  put 

mc,(T2-Tinjp{V2-Vi)  =  mCp{T2-Ti). 

To  simplify  this  equation    eliminate  the  p  and  the  F's  by 
means  of  pV=mRT  from  which  we  obtain 

p  Vi  =  niR  Ti     and     p  V2  =  mR  T2 

for  p  by  hvpothesis  remains  constant. 

Thus  mCv{T2—  T{)-\-jmR{T2  —  Ti)  =  mCp(T2—  Ti) 

R 

or  Cp—Cv  =  ^. 

This  proof  of  the  fundamental  relation  between  Cp  and  c» 
is  based  upon  the  assumption  that  Cp  and  Cv  are  constant 
(as  they  would  be  for  ideal  gases)  or  that  Cp  and  Cv  are 
the  mean  specific  heats. 

By  writing  the  energy  equation  in  its  differential  form 
it  is  readily  shown  that  the  relation  also  holds  for  the 
instantaneous  values  of  varying  specific  heats. 

Thus  dQp  =  dKp+dWp 

^dK,-{-dWp 

or  mcpdT  =  mcvdT-{-—pd  V. 


26 

THERMODYNAMICS 

From 

pV=mRT  with  p  constant 

we  have 

pdV^mRdT, 

so  that 

mcpdT  —  mcvdT^—mRdT, 

whence 

R 

Cp     Cp  —  _ 

as  before. 

The  ratio  —  occurs  frequently  in  thermodynamics.     This 


C-j) 


ratio  will  be  denoted  by  k.    Thus,  by  definition, 


As  Cp  and  Cv  are  constants,  k  is  constant  for  ideal  gases. 
Actually  k  varies  with  the  temperature. 

Specific  Heats  at  Constant  Pressure. — By  means  of 
the  relation  between  Cp  and  Cv  just  established  Cp  may  be 
computed  when  Cv  is  known. 


As 


R 

■C.-J 


fJLCp  — JJLCv'- 


fxR 

''  J ' 


Substituting  the  value  of  R  in  terms  of  /x 

1540 


R  = 


fJ- 


we  find 


ixcp  =  fjLCv-\ — J—  =  iJLCv-\- 1 .98, 


Assuming  Langen's  values  for  /xCv  we  have  for  diatomic 

gases 

)UCp=  6.6i+o.ooo588r. 


THE    LAWS    OF   IDEAL   GASES 


27 


In  general 
and 


Cn  —  Cfi\ 


1.98 


1.98 


For  purposes  of  engineering  calculations  Cp  for  diatomic 
gases  between  32  and  400°  F  is  usually  considered  con- 
stant. The  following  values  may  be  used  for  Cp,  Cc, 
and  k. 


cp 

Ct 

k 

Air 

0. 240 
3 -40 
0.217 
0.244 
0.250 

0.  21 
0-53 

0.171 
2.42 

0155 
0.174 
0.179 

0.16 
0.41 

1 .40 

H2 

1.40 

O2 

1 .40 

N2 

1 .40 

CO 

1.40 

CO2 

1 .31 

NH3 

1 .29 

Exercise    17.  Compute   Mfc,   MCp,   and   k   for   diatomic    gases 
(a)  at32°F,  (6)  at  212°  F. 

Exercise  18.  Show  that  for  diatomic  gases 

257  7 

Cp— Cr=-,    Ci,=~,    Cp=— ,    and    k=— =  1.40 
M  M  M  5 

approximately. 
Exercise  19.  Show  that 


Cp  = 


R     k 

J  k-i 


rr=7^-— ;    k-- 
J  k—i 


Cp 


Exercise  20.  Show  that  the  external  work,  performed  per 
degree  change  in  temperature  when  an  ideal  gas  expands  at  con- 
stant pressure  is  mR. 


28  THERMODYNAMICS 

Exercise  21.  Compute  the  mechanical  equivalent  of  heat 
from  the  following  experimental  data.  One  pound  of  air  at 
n.t.p.  occupied  12.39  cubic  feet,  Cp  for  air  at  32°  F  equals  0.2375 
and  Ce  at  32°  F.  equals  0.1689. 

Exercise  22.  Ten  pounds  of  air  are  heated  at  a  constant 
pressure  of  200  pounds  per  square  inch  absolute  from  50  to 
60°  F. 

(a)  How  much  heat  energy  has  been  supplied? 

(b)  How  much  of  this  energy  is  stored  in  the  air? 
Exercise  23.  One  pound  of   air  at  100°  F  expands  at  con- 
stant pressure  while  20  B.t.u.  are  supplied. 

(a)  What  is  the  final  temperature  of  this  air? 

(b)  How  much  external  work  has  been  done? 

Exercise  24.  How  much  heat  is  required  to  expand  20  cubic 
feet  of  air  to  30  cubic  feet  under  a  constant  pressure  of  60  pounds 
per  square  inch  absolute? 

Exercise  25.  Three  pounds  of  air  confined  under  constant 
pressure  are  heated  from  50  to  100°  F. 

(a)  How  much  heat  is  supplied? 

(b)  What  increase  in  internal  energy  results? 

(c)  How  much  external  work  is  done? 

Section  V 

THE  FUNDAMENTAL   LAWS   FOR   IDEAL   GASES 

The  fundamental  laws  for  ideal  gases  may  be  summed 
up  in  mathematical  language  as  follows. 

pV  =  mRT 

1540 


where  R 


dQ  =  dK+dW, 


THE    LAWS    OF    IDEAL   GASES  29 

and  dK  =  mcAT 

under  all  conditions. 

Ri.pS 

Cn       Ce  —  _  — 

J  M 

As  dW  =  jpdV 

the  energy  equation  may  be  written 

dQ=mc„dT+|pdV. 

This  equation  may  also  be  expressed  in   terms  of  dp, 

instead  of  dV    by  means  of  pV  =  mRT,   from    which   we 

obtain 

pdV-\-Vdp^mRdT 

or  pdV  =  mRdT-Vdp. 

A      A  ^ 

And  as  Cp  — G  =  -y, 

this  equation  reduces  on  elimination  of  R  to 
pdV  =  mJ  {cp  -c,)dT-  Vdp, 

so  that  dQ  =  mCpdT  -  TrVdp. 

Exercise  26.  Show  that  dQ  may  be  expressed  as  follows: 

dV 
dQ=mCf4T+m{cp  —  Cv)  ^~^, 

dQ = mcpdT— m  {cp — Co)  T — , 
P 

dp  dV 

dQ = mcvT \-mcpT-rT-. 

^  p  V 


CHAPTER  II 
CHANGES  OF  STATE  OF  IDEAL   GASES 

Section  VI 
GENERAL   DISCUSSION 

Ideal  gases  obey  the  law 

pv  =  RT, 

where  i?  is  a  constant  for  any  given  gas. 

The  state  of  a  gas  is  determined  by  its  volume,  its  pres- 
sure, and  its  temperature. 

In  this  chapter  will  be  discussed  some  changes  of  state 
which  a  gas  undergoes  when  any  or  all  of  its  conditions 
of  volume,  pressure,  and  temperature  vary. 

Characteristic  Equation  and  Surface. — As  the  state  of 

a  gas  depends   upon  three  conditions  namely  p,  v,  and  T 

this  state  may  be  represented  graphically  by  a  point  in 

space   referred   to   three   rectangular   axes.     The  locus   of 

all  points  representing  possible  states  of  the  gas  is  a  surface 

whose  equation  is 

pv=RT, 

the  characteristic  equation  of  the  ideal  gas. 

Every  gas  has  its  own  characteristic  equation  and  corre- 
sponding surface  because  the  parameter  R  of  the  equation 
depends  upon  the  nature  of  the  gas, 

30 


CHANGES  OF  STATE  OF  IDEAL  GASES 


31 


The  characteristic  surface  of  any  ideal  gas  is  a  hyperbolic 
paraboloid,  a  portion  of  which  is  shown  in  Fig.  2.  That 
this  is  so  is  evident  from  the  following  analysis  of  the 
equation  pv  =  RT. 

(i)  The  equations  of  all  sections  of  the  surface  made 


Fig.  2. 


by  planes  parallel  to  the  /Ji'-plane,  for  which   T=Ti,   a 
constant,  are 

pv  =  RTi  =  a,  constant. 

These  curves  are  thus  equilateral  hyperbolas. 

(2)  The  equations  of  all  sections  of  the  surface  made 
by  planes  parallel  to  the  pT-plane,  for  which  v  =  vi,  a 
constant,  are 


32  THERMODYNAMICS 

These    sections   are    thus    straight    lines  passing  thru  the 
D-axis. 

(3)  The  equations  of  all  sections  of  the  surface  made 
by  planes  parallel  to  the  z^r-plane,  for  which  p  =  p\,  a 
constant,  are 

These  sections  are    thus    straight    lines  passing    thru  the 
^-axis. 

The  points  on  any  line  drawn  upon  the  characteristic  sur- 
face of  an  ideal  gas  between  the  points  {pi,  vi,  Ti)  and 
{p2,  V2,  T2)  represent  a  possible  sequence  of  values  thru 
which  p,  V,  and  T  may  pass  while  one  pound  of  the  gas 
changes  its  state  from  a  condition  determined  by  pi,  vi, 
and  Ti  to  another  condition  determined  by  p2,  vo,  and  T2. 

Exercise  27.  What  condition  is  imposed  upon  a  change  of 
state  following  the  line  (a)  AOE,  (b)  HOD,  (c)  FOB,  (d)  BC, 
(e)  HA ,  (/)  GF  traced  upon  the  characteristic  surface  of  an  ideal 
gas,  Fig.  2? 

Exercise  28.  How  does  {a)  an  increase  in  pressure  at  constant 
volume  {b)  a  decrease  in  volume  at  constant  pressure  affect 
the  temperature  of  an  ideal  gas? 

Solve  analytically,  and  also  graphically  by  means  of  Fig.  2. 

Instead  of  using  lines  traced  upon  the  characteristic  sur- 
face to  represent  changes  of  state  of  a  gas  it  is  customary 
to  use  only  the  projections  of  these  lines  upon  the  coor- 
dinate planes.     As  the  external  work  done  by  a  gas  during 

a  change  of  state  always  equals    \  pdV  and  as  this  external 

work  is  of  great   importance  in    engineering    applications 
the  projection  upon  the  />z)-plane  is  the  most  useful  pro- 


CHANGES    OF    STATE    OF    IDEAL    GASES  33 

jection.  Fig.  2  shows  the  pv-curves  representing  some  of 
the  more  important  changes  of  state.  The  projected  lines 
naturally  show  only  the  changes  which  occur  in  the  pres- 
sure and  the  volume  of  the  gas;  no  changes  in  temperature 
can  here  be  indicated. 

Important  Changes  of  State. — It  is  evident  that  changes 
of  state  may  occur  in  innumerable  ways.  Any  line  drawn 
upon  the  characteristic  surface  would  represent  a  possible 
way  in  which  a  change  of  state  might  occur.  Of  all  pos- 
sible changes  of  state  the  only  ones  of  importance  to  the 
engineer  are: 
Isothermal  changes,  during  which  the  temperature  remains 

constant, 
Isometric  changes,  during  which  the  volume  remains  con- 
stant, 
Isopiestic  changes,  during  which  the  pressure  remains  con- 
stant, 
Isodynamic  changes,  during  which  the  internal  energ}   re- 
mains constant, 
Adiabatic  changes,  during  which  no  heat  is  received  from 

or  rejected  to  external  bodies, 
Polytropic  changes,  during  which  the  heat  supplied  to  or 
withdrawn  from  the  gas  by  external  bodies  is  directly 
proportional  to  the  change  in  temperature. 
The  discussion  of  any  change  of  state  should  include: 
(i)  The    equations    showing    the    relations    between    p, 
V,  and  T  during  the  change, 

(2)  The  external  work  performed  by  or  on  the  gas  during 
the  change, 

(3)  The  quantity  of  heat  supplied  to  or  rejected  by  the 
gas  during  the  change, 


34  THERMODYNAMICS 

(4)  The  change  in  the  intrinsic  or  internal  energy  of  the 
gas  during  the  change  of  state. 

It  should  be  remembered  that  the  internal  or  intrinsic 
energy  of  an  ideal  gas  changes  only  with  the  temperature 
and  that  this  change  in  internal  energy  during  any  change 
of  state  always  equals 

mCvdT = mcJT2—  Ti). 

Ti 

Section  VII 

ISOTHERMAL,     ISOMETRIC,    ISOPIESTIC,     AND 
ISODYNAMIC    CHANGES 

Isothermal  Change  of  State. — The  equations  for  this 
change  of  state  involve  first 

pv  =  RT 

the  fundamental  equation  which  is  always  true.     In  addition 
we  impose  the  condition  of  constant  temperature  so  that 

T=T\,  a  constant. 

Thus  pv  =  RTx  =  C\,  a  constant. 

This  is  Boyle's  law  and  it  is  the  equation  of  the  ^z'-curve 
for  isothermal  change  of  state. 

Exercise  29.  What  is  the  value  of  d  for  the  isothermal 
change  of  state  of  one  pound  of  air  at  90°  F  ? 

Exercise  30.  Show  that  the  construction  illustrated  in  Fig. 
3  yields  a  curve  for  which  pv  =  piVi  =  a.  constant. 

Exercise  31.  Draw  to  scale  the  isothermal  ^I'-curve  for 
one  pound  of  air  at  (a)  90°  F,  (b)  200°  F. 


CHANGES   OF   STATE   OF   IDEAL   GASES 


35 


The  external  work  performed  is  represented  graphically 
by  the  area  under  the  pv-cnrve,  Fig.  3,  for 


AW 


I  r^' 


The  sign  of  AW  will  be  assumed  positive  when  external 
work  is  done  by  the  gas  during  expansion  and  negative 
when  external  work  is  done  on  the  gas  during  compression. 


Fig.  3. 


Exercise  32.  Show  that  the  external  work  performed  during 
the  isothermal  change  of  state  of  one  pound  of  ideal  gas  from 
/>i.  t'l,  Ti  to  p,  V,  Ti  is 

AW=^PiV,  lege  (^j, 

when  measured  in  B.t.u.,  and  also  equals 

jM:l0ge(^ 


-RT^  loge 
J  \p 


where 


loge(x)  =2.303  logio(x). 


The  heat  absorbed  (or  rejected)  by  a  gas  during  isothermal 
change  of  state  cannot  be  computed  by  means  of  a  specific 
heat,  for  the  change  in  temperature  is  zero. 


36  THERMODYNAMICS 

We  have,  however,  the  equation 

AQ=AK-\-AW. 
Here  during  isothermal  change 

AA'  =  o, 

I  /  V 

so  that  AQT  =  AWT  =  'ypivi  loge    — 

Exercise  t,^.  How  much  heat  must  be  supplied  to  3  pounds 
of  air  expanding  at  a  constant  temperature  of  100°  F  from 
200  to  30  pounds  per  square  inch  absolute? 

Exercise  34.  Gas  expands,  no  heat  is  suppHed  to  it.  Where 
does  the  energy  necessary  for  the  performance  of  external  work 
come  from?     Is  this  an  isothermal  change  of  state? 

Exercise  35.  During  isothermal  expansion  all  heat  suppliea 
is  transformed  into  external  work.  Why  is  this  not  an  ideal 
way  of  transforming  heat  into  mechanical  energy? 

Isometric  Change  of  State.— For  all  changes  of  state 
we  have,  pv  =  RT,  if  we  now  impose  upon  this  general 
law  change  at  constant  volume  then 

z'  =  z;i  =  a  constant. 

Combining  these  equations  we  find 

^  =  — =  C2,  a  constant. 

This  is  one  form  of  the  law  of  Charles. 

Exercise  36.  Interpret  the  above  equations  graphically  by 
means  of  Fig.  2. 

Exercise  37.  Describe  the  pv-curve  for  isometric  change  of 
state.  How  much  external  work  is  done  during  this  change 
of  state? 


CHANGES  OF  STATE  OF  IDEAL  GASES         37 

Exercise  38.  How  much  heat  must  be  supplied  to  a  gas 
(specific  heat  at  constant  volume,  c»)  in  order  to  increase  its 
pressure  from  pi  to  p2  pounds  per  square  inch  absolute,  the 
volume  remaining  Vi  during  the  change? 

What  becomes  of  this  heat? 

Exercise  39.  Compute  C2  for  air  under  a  pressure  of  200 
pounds  per  square  inch  absolute  and  at  90°  F.  For  what  volume 
of  air  does  this  constant  apply? 

Isopiestic  Change  of  State. — 

Exercise  40.  Write  the  equations  governing  isopiestic  change 
of  state.     Interpret  these  equations  on  Fig.  2. 

Exercise  41.  Describe  the  ^z^-curve  for  isopiestic  change  of 
state  and  compute  the  external  work  done  during  a  change 
from  pi,  Vi,  Ti  to  pi,  v,  T. 

Exercise  42.  Is  the  internal  energy  of  a  gas  increased  or 
diminished;  is  external  work  done  by.  or  on;  must  heat  be 
supplied  to,  or  withdrawn  from  a  gas  during  isopiestic  (a)  expan- 
sion, (b)  compression? 

Exercise  43.  Ten  pounds  of  air  are  kept  under  a  constant 
pressure  of  200  pounds  per  square  inch  absolute. 

(a)  What  temperature  has  this  air  when  it  occupies  12  cubic 
feet? 

(b)  If  its  initial  temperature  was  300°  F  and  its  volume  is 
then  reduced  to  12  cubic  feet  what  external  work  was  done 
upon  the  gas? 

(c)  How  much  intrinsic  energy  did  it  lose? 

(d)  How  much  heat  was  withdrawn  from  the  gas? 

Isodynamic  Change  of  State. — By  definition  during  an 
isodynamic  change  of  state  the  internal  energy  remains 
constant. 

The  internal  energy  of  an  ideal  gas  is  a  function  of  the 


38  THERMODYNAMICS 

temperature  only.     Therefore  constant  internal  energy  im- 
plies constant  temperature. 

Thus  isodynamic  changes  of  state  of  ideal  gases  are 
isothermal  changes  of  state. 

Exercise  44.  Express  the  change  in  internal  energy  of  an 
ideal  gas  in  terms  of  Ct  and  show  from  this  expression  that  an 
isodynamic  change  is  an  isothermal  change. 

Section  VIII 
ADIABATIC    CHANGE    OF    STATE    OF   IDEAL    GASES 

Definition  and  General  Discussion. — During  an  adia- 
batic  change  of  state  no  heat  is  received  from  or  rejected 
to  external  bodies.  Such  a  change  of  state  would  occur 
if  the  gas  could  be  insulated,  as  far  as  heat  is  concerned, 
from  all  surrounding  bodies.  In  practice  adiabatic  changes 
are  approximated  when  expansion  or  compression  takes 
place  so  rapidly  that  little  time  is  allowed  for  the  transfer 
of  heat. 

Gas  expanding  adiabatically  does  external  work.  The 
energy  thus  converted  into  external  work  is  not  derived 
from  an  external  source  of  supply;  the  gas  is  insulated. 
This  energy'  must  therefore  be  drawn  from  the  internal 
(intrinsic)  energy  of  the  gas.  As  the  internal  energy  of 
an  ideal  gas  depends  only  on  its  temperature,  the  tem- 
perature of  an  adiabatically  expanding  gas  must  fall.  The 
internal  energy  which  disappears  during  adiabatic  expan- 
sion reappears  as  external  work  performed  during  the 
expansion. 

In  Fig.  2  the  line  GOC  approximately  represents  an 
adiabatic  change  of  state.     Note  the  drop  in  temperature 


CHANGES    OF    STATE    OF   IDEAL   GASES  39 

of  the  gas  during  expansion  and  the  rise  in  temperature 
during  compression. 

Exercise  45.  Explain  the  fact  that  the  temperature  remains 
constant  during  isothermal  expansion  and  yet  external  work  is 
done. 

Is  an  mcrease  in  temperature  possible  during  an  expansion 
of  a  gas?     Explain  by  means  of  Fig.  2. 

Exercise  46.  What  terms  in  the  equation 

disappear  during  an  adiabatic  change  of  state  of  an  ideal  gas? 
Why  are  these  terms  equal  to  zero? 

Does  a  gas  possess  more  or  less  internal  energy  after  adiabatic 
expansion?     To  what  is  the  change  in  internal  energy  equal? 

The  Equations  Governing  Adiabatic  Change  of  State. 
The  fundamental  equation 

pv  =  RT (i) 

must  always  be  satisfied. 

During  adiabatic  change  we  have  in  addition  the  con- 
dition that 

dQ  =  o (2) 

This  last  equation  must  be  exiiressed  in  terms  of  p,  v, 
and  T  to  make  it  useful. 

As  dQ  =  mc,dT-\-jpdV 

we  have  CvdT-\-—pdv  =  o (3) 

a  differential  equation  representing  the  conditions  of  adia- 
batic change  of  state  in  terms  of  p,  v,  and  T. 


40  THERMODYNAMICS 

In  order  to  integrate  this  equation  it  must  be  expressed 
in  terms  of  two  variables  by  means  of  the  fundamental 
relation  pv  =  RT. 

We  may  eliminate  p  from  equation  (3)  by  substituting  for 

,  RT    , 
p  its  equal  — ,  thus 

CvdT-\ — =  —dv  =  o. 
J  V 

But  -J  always  equals  Cp—Cv,  so  that 

T 

CvdT-\-  (cp—Cv)—dv  =  o, 

V 

dT         (cp       \  dv 
or  -=r=  — I— —  il— . 

T  \cv       /  V 

For  ideal  gases  Cp  and  Cv  are  constant,  so  that 

a  constant,  by  definition,  and  the  differential  equation  may 

be  written 

dT         ,,       .dv 

Integrating  between  the  initial  state  p\,  vi,  T\,  and 
the  final  state,  p,  v,  T  we  have 


logerl^=-(^-i)loge  4 


log«jr=-(^-i)loge^^ 


CHANGES 

OF 

STATE    OF 

IDEAL 

GASES 

log* 

T 

'Ti" 

=  log< 

^(^ 

-ft 

T 

\V 

y-i 
/ 

41 


whence 


This  equation  together  with  pv  =  RT  definitely  fixes  the 
relations  between  p,  v,  and  T  during  an  adiabatic  change 
of  state. 

Exercise  47.  (a)  Deduce,  in  the  manner  outlined  above,  an 
equation  in  terms  of  p  and  v  from  dQ  =  o. 
(h)  Same  in  terms  of  p  and  T. 

The  equations  governing  the  adiabatic  change  of  state 
of  any  ideal  gas  are 

pv  =  RT 

together  with  any  one  of  the  following  equations 

k-l 


ft-i 

Pl_ 

/v^ 

k 

Ti 

/Pi^ 

k 

Ti 

p 

\Vl/ 

} 

T 

Vp. 

) 

T 

\Vl/ 


Exercise  48.  Interpret  these  equations  geometrically  by 
means  of  Fig.  2. 

Exercise  49.  Deduce  the  last  three  equations  by  means  of 
the  equations  given  in  Exercise  26. 

Starting  with  pv  =  RT  and  pi^^pivi'^,  the  other  rela- 
tions between  p,  v,  and  T  existing  during  adiabatic  change 
of  state  may  be  deduced  by  eliminating  either  p  or  ;;.  As 
an  example  let  us  eliminate  p. 

From                              pv^  =  p\V\^ 
we  have  — =(  — I (i) 


42  THERMODYNAMICS 

Also  pv  =  RT 

pV  p\Vx  ,    . 

gives  ^  =  R  =  ^ (2j 

To  eliminate  p  we  have  from  equation  (2) 

p  _viT 
pi     vTi 

and  from  equation  (i)       7~~  ("~)  ' 


therefore 


or 


Exercise  50.  Starting  with  pv  =  RT  and  pv^=piv^  deduce 
the  relation  between  p  and  T  for  an  adiabatic  change  of  state. 

Exercise  51.  Show  that  the  final  temperature  of  an  ideal 
gas  after  an  adiabatic  change  of  state  depends  only  upon  the 
initial  temperature  and  upon  the  ratio  of  expansion  (or  upon 
the  ratio  of  the  initial  to  the  final  pressure)  but  not  upon  the 
individual  volumes  (or  pressures)  involved. 

Exponential  and  Logarithmic  Computations. — As  the 
equations  just  developed  show,  thermodynamic  computa- 
tions often  require  logarithmic  computations.  The  follow- 
ing examples  will  serve  as  a  review  and  they  should  be 
carefully  studied. 

I.  What  does  (0.035)"'^^  equal? 

Let  X=(o.o35)''-'' 

then  logio  A' =  (.21)  logio  (.035) 

=  (.2i)(8.5435-io) 


CHANGES    OF    STATE    OF   IDEAL    GASES  43 

=  (.2l)(- 1.4565) 
=  -0.3059 
=  10- (0.3059) -10 
=  9.6941  —  10 

therefore  "  N'  =  0.4944 

2.  Find  the  value  of  loge  (0.085), 
where  6=2.718  .  .  . 

Let  x=loge  (0.085) 

then  6^  =  0.085. 

Taking  logarithms  to  the  base  10  for  which  complete  tables 
are  at  hand  we  have 

xlogio  e  =  logio  (.085), 

or  x  =  : logio  (.085) 

logio  e 

=  -^{8.9285-10} 
•434 

=  2.303{-i.o7i5l 

=  -2.468. 
Note  that  in  general 

loge  N  =  . logio  .¥=2.303  logio  N. 

logio  « 

Exercise  52.  One  pound  of  air  expands  adiabatically  from 
300  pounds  per  square  inch  absolute  and  200°  F  to  15  pounds 
per  square  inch  absolute. 

(a)  What  is  its  final  temperature? 

(b)  What  is  its  initial  volume? 

(c)  What  is  its  final  volume? 


44  THERMODYNAMICS 

Exercise  53.  Four  cubic  feet  of  air  at  60°  F  are  compressed 
adiabatically  until  the  volume  is  reduced  to  ^  cubic  foot. 

(a)  What  is  its  final  temperature? 

(b)  What  relation  exists  between  the  final  and  the  initial 
pressures? 

(c)  If  the  mass  of  air  had  been  one  pound  what  would  have 
been  the  final  pressure? 

The  External  Work  Performed  by  or  on  an  ideal  gas 
during  an  adiabatic  change  of  state  may  be  computed  in 
two  ways, 

(i)  by  means  of  the  area  under  the  pv-cuvve  repre- 
senting adiabatic  change.  That  is,  the  area  on  the  pv- 
plane  bounded  by  the  curve  pv^  =  pivi^,  the  z'-axis,  and 
the  ordinates  of  the  points  representing  the  initial  and  the 
final  states  of  the  gas. 

(2)  by  means  of  the  change  in  internal  energy  which 
occurs  during  the  adiabatic  change  of  state,  for  all  external 
work  is  done  at  the  expense  of  the  internal  energy  of  the 
gas.  Thus  any  loss  in  internal  energy  equals  the  external 
work  performed,  or 

AW=  -AK=  -mc„(T-Ti), 

where  the  result  is  expressed  in  B.t.u. 

Exercise  54.  Show  that  the  external  work  performed  during 
an  adiabatic  change  of  state  when  expressed  in  foot-pounds 
equals 

JmcvTi  (  I  —  :^  )  =  JmCvTi  \i—(-^]       \=  JmcTi  |  i  —  (  T" 

where  JtncvTi  may  be  replaced  by 

JcvpiVi  piVi 


CHANGES  OF  STATE  OT  IDEAL  GASES         45 

Exercise  55.  Show  by  means  of  the  pv-curve  that  the  external 

work  done  during  an  adiabatic  change  of  state   (expressed  in 

foot-pounds)  equals 

p.Vi  -pV 

k-i    ■ 

Exercise  56.  Prove  that  the  expression  derived  in  Exercise 
55  equals  the  expressions  derived  in  Exercise  54. 

Exercise  57.  Compute  the  work  done  during  the  expansion 
of  the  air  discussed  in  Exercise  52 

(a)  by  means  of  the  results  of  Exercise  54, 

(b)  by  means  of  the  results  of  Exercise  55. 

Exercise  58.  Compute  the  work  for  Exercise  53  assuming 
the  mass  of  air  to  be  one  pound. 

The  Heat  Absorbed  from  or  rejected  to  external  bodies 
during  an  adiabatic  change  of  state  is  by  definition  equal 
to  zero. 

The  Change  in  Internal  Energy  during  an  adiabatic 
change  of  state  is 

AK  —  7nCviT—Ti). 

Section  IX 

POLYTROPIC  CHANGE  OF  STATE  OF  IDEAL  GASES 

Definition  and  General  Discussion. — During  an  isopiestic 
expansion  from  the  state  determined  by  pi,  vi,  Ti  to  the 
state  determined  by  p\,  V2,  To  the  heat  supplied  to  an 
ideal  gas  must  be  suiBcient  to  increase  the  internal  energy 
and  to  perform  the  external  work  done  by  the  expanding 
gas.     Thus  the  heat  supplied  must  equal 

^Qp^mcv{T2—Ti)-\-m~.n{v2  —  Vi) 


=^mU+^{T2-Ti). 


46  THERMODYNAMICS 

Here   the   factor    (ce+-y)    is   evidently   a   specific   heat. 

It  equals  c^  as  shown  on  page  25. 

During  any  change  of  state  of  an  ideal  gas  the  specific 
heat  c  defined  by  the  equation 

dQ  =  mcdT 

may  similarly  be  regarded  as  composed  of  two  terms, 
one  always  being  Cy  and  the  other  term  being  analogous  to 

—  but  not  equal  to  it.     This  second   term    depends  upon 

the  manner  in  which  the  gas  changes  its  state  and  upon 
the  external  work  done  during  this  change.  Therefore 
the  specific  heat  of  a  given  gas,  c,  may  have  as  many 
different  values  as  there  are  ways  in  which  the  gas  may 
change  its  state.  Even  tho  c„  be  considered  constant 
c  need  not  remain  constant  during  all  possible  changes 
of  state. 

By  definition,  a  polytropic  change  of  state  is  any  change 
of  state  during  which  the  specific  heat  c  does  remain  con- 
stant, see  page  33. 

We  must  now  determine  the  relations  between  p,  v, 
and  T  during  polytropic  changes  of  state. 

The  Equations  Governing  Polytropic  Change  of  State 

of  Ideal  Gases. — ^The  equation  governing  all  changes  of 

state  of  ideal  gases  is 

pv  =  RT. 

In  order  to  limit  the  change  to  a  polytropic  one  the  con- 
ditions expressed  by  the  equation 

dQ  —  mcdT, 

where  c  is  constant,  must  also  be  satisfied. 


CHANGES   OF   STATE   OF   IDEAL   GASES  47 

This  second  equation  of  condition  must  now  be  expressed 
in  terms  of  p,  v,  and  T.     As 

dQ^mc4T-\-jpdV 
under  all  conditions  of  change  of  state  we  may  write 
cdT  =  CvdT-\-  -jpdv, 

or  (c-Cv)dT  =  —pdv. 

To  establish  a  relation  between  the  p's  and  the  v's  T 
must  be  eliminated  by  means  of  pv  =  RT,  from  which  we 
obtain 

dT=^(pdv-\-vdp) 
K 

SO  that  {c—Cv){pdv-\-vdp)  =  —pdv. 

Replacing  —  by  Cp  —  Cv  and  regrouping  the  terms  we  have 

{c—Cp)pdv-\-  {c—Cv)vdp  =  o. 

Integrating    between    the    limits  pi,  I'l,  Ty  and  p,  v,  T 
we  have 

{c-Cp)  log  ^+{c-Cv)  log  —  =  o, 
Vi  px 

y  '\-p)  ■ 

Comparing   this   equation   with   the  analogous  equation 
for  adiabatic  change  of  state,  namely, 

P      W' 


48  THERMODYNAMICS 

we  note  that  the  corresponding  polytropic  equation  may  be 
written 

C-Cp 

P      \n, 
Here is  a  constant,  for  Cp  and  c»  are  constants  for 

C~Cv 

ideal  gases  and  c  is  a  constant  by  definition  for  polytropic 
changes  of  state.  Let  this  constant  exponent  be  denoted 
by  n,  then 

P     \n, 

Thus   during   any   polytropic   change   of   state   of    ideal 

gases,  we  have 

pv  =  RT 

and  pv"  =  piVi", 

where  the  constant  n  has  the  value 


C      Of) 

n  = 


Solving   the   last    equation   for   c   we    may   express   the 

specific   heat   for   any   given   polytropic   change    in    terms 

of  n  as  follows: 

n-k 

n— I 

Why  do  not  the  above  equations  hold  for  an  actual  gas? 

Exercise   59.  Deduce  the  relation  between  v  and   T  for  a 
polytropic  change  of  state  of  an  ideal  gas  starting  with 
(c)  the  fundamental  equation  dQ=mcdT, 
(jb)  the  equations  pv=RT  and  pv^^piVi**. 


CHANGES    OF    STATE    OF   IDEAL   GASES  49 

Exercise  60.  Deduce  the  relation  between  p  and   T  for  a 
polytropic  change  of  state,  starting  with 
{a)  the  fundamental  equation  dQ  =  nicdT, 
(b)  the  equations  pv  =  RT  and  pv^  =  piv^. 
Exercise  61.  Deduce  the  equations 

n-l 

T      /v,\"-i       T      /p 


pv'=pivr;  .     .      ,  , 

JLi      \v/  li      \pi 

expressing  the  conditions  involved  in  polytropic  changes  of 
slate  of  ideal  gases  from  the  equations  of  Exercise  26. 

Discussion  of  the  Poljrtropic  Equations.  — The    equa- 
tions to  be  discussed  are 

K=/'r^'i" (i) 

where  n  may  have  any  constant  value,  and 

n-k 

C  = Cv (2) 

n—  I 

from  which  the  value  of  the  constant  specific  heat  may 
be  computed  for  any  given  value  of  n. 
(i)  When  71  =  o,  equation  (i)  becomes 

P  =  Pi 
and  from  equation  (2) 

C  —  RCi)  —  Cp, 

Therefore  a  polytropic  change  during  which  n  =  o  is  an 
isopiestic  change  of  state,  represented  in  Fig.  4  by  the 
line  aoe. 

(2)  When  n=i,  equation  (i)  becomes 

pv  =  piVi 

and  from  equation  (2) 

i-k 

c= Ct=00  . 

o 


50 


THERMODYNAMICS 


Therefore  a  polytropic  change    during  which  n=i    is  an 
isothermal  change,  represented  in  Fig.  4  by  the  line  bof. 

Note  that  the  specific  heat  of  the  gas  is  now  infinite. 
The  gas  thus  has  an  infinite  capacity  for  heat;  it  can 
absorb  or  reject  any  quantity  of  heat  without  change 
in  temperature  provided  it  obeys  the  laws  pv  =  RT  and 
pv=pivi.  During  isothermal  expansion  this  actually  occurs. 
We  must  however  bear  in  mind  that  the  heat  so  absorbed 
without   change   in   temperature   does   not   remain   in    the 


P 


f,g    h 


g  n-o ;  c^C] 


Fig.  4. 

gas  but  is  at  once  transformed  into  external  work  performed 
by  the  gas,  and  the  heat  rejected  without  change  in  temper- 
atiu-e  is  not  derived  from  the  gas  but  from  the  external 
work  performed  simultaneously  on  the  gas,  the  internal 
energy  of  the  gas  remaining  constant  thruout  the  change. 
(3)  When  n  =  k,  equation  (i)  becomes 

and  equation  (2)  shows  that 


c  =  o. 


CHANGES  OF  STATE  OF  IDEAL  GASES         51 

Therefore    a    polytropic  change  during  which  n  =  k  is  an 
adiabatic  change,  represented  in  Fig.  4  by  the  Hne  cog. 

Note  that  when  pv*^  —  pivi^  the  gas  has  absolutely  no 
capacity  for  heat  (c  =  o).  As  dQ  =  mcdT  and  c  =  o,  an 
infinitesimal  quantity  of  heat  supplied  to  the  gas  under 
these  conditions  would  cause  an  infinite  rise  in  temperature. 
This  must  be  so  for  under  adiabatic  conditions  no  heat 
should  be  added  to  or  be  extracted  from  the  gas. 

(4)  When  n=  Qc  ,  let  n  =  -,  where  a  and  b  are  constants, 
0 

then  equation  (i)  becomes 

and  as  b  must  approach  o  as  n  approaches   00    we   have 
for  n=  00 

From  equation  (2)  we  have 

--k 

b  a-bk 

C  —  — — ^—  ^p  —  ,    Cf)» 

a  a  —  b 

Thus  when  n=<x> ,  b^^o  and 

C=Cv. 

Therefore  a  polytropic  change  during  which  n=  00  is 
an  isometric  change,  represented  in  Fig.  4  by  the  line  doh. 

Polytropic  changes  of  state  thus  include  all  the  changes 
of  state  previously  considered,  together  with  other  changes 
of  state.     It  should  however  be  noted  that  the  polytropic 


52  THERMODYNAMICS 

conditions  do  not  include  all  possible  changes  of  state 
but  only  those  during  which  the  specific  heat  remains 
constant. 

Of  all  the  polytropic  changes  of  state  other  than  the 
isopiestic,  isometric,  isothermal,  and  adiabatic  the  only 
polytropics  used  in  engineering  are  the  ones  for  which 
n  lies  between  i  and  k.  The  ^z'-curves  of  these  polytropics 
lie  between  the  lines  fob  and  goc  in  Fig.  4.  For  these 
polytropics  the  specific  heat  is  negative. 

As  dQ=mcdT  it  follows  that  when  c  is  negative  and 
dQ  is  positive,  dT  must  be  negative  therefore  when  heat 
is  supplied  a  drop  in  temperature  occurs.  This  appar- 
ently inconsistent  condition  will  be  understood  when  it 
is  remembered  that  the  external  work  performed  plays  an 
important  part  in  the  discussion.  Thus  under  the  condition 
pv'^  =  p\Vi^  when  n  lies  between  i  and  k  work  is  performed 
so  rapidly  during  the  expansion  of  the  gas  that  not  only 
the  whole  heat  supplied  is  converted  into  external  work 
but  this  heat  does  not  suffice  to  perform  the  work  and  the 
store  of  internal  energy  of  the  gas  is  called  upon  to  aid  with 
the  evident  result  that  a  drop  in  temperature  occurs. 

During  compression  under  these  conditions  the  work 
done  on  a  gas  and  converted  into  heat  cannot  be  com- 
pletely rejected  by  the  gas  and  that  which  remains  in  the 
gas  as  an  increase  in  internal  energy  causes  a  rise  in  tem- 
perature even  tho  heat  is  removed  from  the  gas.  Thus 
altho  dQ  is  negative,  JT  is  positive  because  c  is  negative. 

Exercise  62.  Trace  the  curve  pv^=piVi^,  assuming  /'i  =  3 
and  fi  =  2,  by  means  of  seveial  points,  for  (a)  n=  —  i,  {b)  11=  —\, 
(c)  n=  —  10.     What  are  the  corresponding  values  of  c? 

Exercise  63.  Describe  the  positions  of  the  lines  on  the  char- 


CHANGES    OF    STATE   OF   IDEAL   GASES  53 

acteristic  surface  (Fig.  2)  corresponding  to  pv'*  =  piVi'*  for  values 
of  n  between 

(a)  +  00  and  i 

(b)  I  and  o 

(c)  o  and  —  =0  , 

for  both  compression  and  expansion. 

Exercise  64.  Write  the  equations  of  the  pv-curves  for  iso- 
thermal, adiabatic,  isopiestic,  and  isometric  changes  of  state. 

(a)  What  values  of  a  and  b  in  //  =  —  and  thus  what  values 

b 

of  n  will   change  the  polytropic  law  pv"'  =  piVi*  into  each  of  the 
above  mentioned  changes  of  state? 

(b)  Discuss  the  value  and  the  physical  meaning  of  the  specific 
heat  for  each  of  the  above  cases. 

The    Signs    of    dQ,   dK,    and    dW. — In   the   equation 

dQ=dK-^dW 
we  have  assumed 
dQ  positive  when  heat  is  supplied  to  the  gas  from  some 

external  source, 
dK  positive  when   the  internal  energy  (and  temperature) 

increases, 
dW  positive  when  external  work  is  done  by  the  gas. 

Evidently  during  any  polytropic  expansion  dW  is 
positive;  during  any  polytropic  compression  dW  is  nega- 
tive. 

From  Fig.  2  it  may  be  seen  that  an  expansion  does  not 
necessarily  entail  a  rise  or  a  fall  in  temperature  with  cor- 
responding increase  or  decrease  in  the  internal  energy. 
For  all  polytropic  expansion  lines  lying  in  front  of  the 
isothermal  OB  (on  the  surface  HOB)  the  temperature 
and    the    internal  energy   increase  during  expansion;  thus 


54  THERMODYNAMICS 

dT  and  dK  are  positive.  For  all  polytropic  expansion 
lines  lying  behind  the  isothermal  OB  (on  the  surface  DOB) 
the  temperature  and  the  internal  energy  decrease  during 
expansion;  thus  dT  and  dK  are  negative. 

To  determine  the  sign  of  dK  for  any  polytropic  change 
of  state  note  that  it  depends  only  upon  the  sign  of  dT. 
To  find  the  sign  oi  dT  analytically  we  may  proceed  as 
follows,  starting  with  the  equations  governing  polytropic 

change, 

pv  =  RT,    pv''=pivi": 

where  p  and  v  are  both  variable,  eliminate  one  of  these 
variables.  If  the  change  in  T  is  to  be  expressed  in  terms 
of  a  change  in  volume  eliminate  p.     Thus 

T     Z'l^'i"  l-n 
T=         v^   ^ 

K 

and  dT=  (^yi-n)(v-){dv). 

The  first  factor  is  always  positive,  the  sign  of  the  second 
depends  upon  the  value  of  n,  the  third  factor  is  always 
positive,  and  the  sign  of  dv  is  positive  for  an  expansion  and 
negative  for  a  compression. 

The  sign  of  dT  may  also  be  found  from  the  equation 

Ti     \v 
To  illustrate,  during  an  expansion  v>vi  so   that  — <i- 


CHANGES  OF  STATE  OF  IDEAL  GASES         55 

If  n  is  greater  than   unity   (  — )         is  less  than    one  and 
T<T\,  therefore  dT  is  negative.      If   however  —  <i  and 

V 

n  is  less  than  unity,  (  —  )        is  greater  than  one  and  T>Ti, 

therefore  dT  is  positive. 
To  determine  the  sign  of  dQ,  remember  that 


dQ  =  nicdT, 

so  that  the  sign  of  dT  does  not  alone  determine  the  sign 
of  dQ.  The  sign  of  c,  the  specific  heat  under  the  particular 
law  pc^'  =  piVi^,  must  also  be  found.  As  an  illustration, 
during  any  expansion  for  which  n  lies  between  i  and  k, 
c  will  be  negative  (see  page  52)  and  dT  will  also  be  neg- 
ative therefore  dQ  must  be  positive.  Under  these  con- 
ditions heat  must  be  supplied  to  the  gas  even  tho  the 
internal  energy  diminishes.  The  external  work  done  ab- 
sorbs not  only  the  loss  of  internal  energy'  but  the  heat 
supplied  as  well. 

If  n  is  greater  than  k,  c  is  positive  and  dT  is  negative, 
therefore  dQ  must  be  negative.  Under  these  conditions 
the  expansion  can  only  proceed  according  to  the  law 
pv^  —  pii'i"  provided  heat  is  withdrawn  from  the  gas.  The 
loss  in  internal  energy  is  so  large  that  all  of  the  energy 
lost  by  the  gas  cannot  be  utilized  in  the  performance  of 
external  work. 

Exercise  65.  (a)  Check  the  statements  in  the  following  table. 
(b)  Sketch  a  figure  similar  to  Fig.  4  and  indicate  the  areas  in 
which  the  polytropics  lie  for  which  (i)  aW  is  +  and  — ,  (2) 
AQ  is  -f-  and  — ,  (3)  aK  is  +  and  — . 


56  THERMODYNAMICS 

During  polytropic  changes  of  state  we  have 


when  the 
values  of  n 
lie  between 

and  the  pv-c\xT\es, 
Fig.  4,  lie 
between 

then  the 
values  of  c 
lie  between 

and  the  signs  of 

dW 

dK 

dQ 

are 

c 
.2 

—  00  and  o 

oh  pnd  oa 

Cv  and  fp;      c  + 

+ 

+ 

+ 

1 

o  and  I 

oa  and  ob 

Cp  and  +a)c;  + 

+ 

+ 

+ 

I  and  k 

oh  and  oc 

—  CO  ando;       c  — 

+ 

— 

+ 

1 

k  and  +  00 

oc  and  od 

oandctj;     c  + 

+ 

— 

— 

c    r 

o 

1 

—  00  and  o 

od  and  oc 

c«andcp;     <:  + 







o 

o  and  I 

oe  and  of 

Cpand  +00  ;  c-\- 

- 

- 

— 

I  and  k 

of  and  og 

—  oo  ando;       c  — 

- 

+ 

- 

o 

k  and  +  00 

og  and  oh 

oandcs;     0  + 

- 

+ 

+ 

""    I- 

The  Experimental  Determination  of  n. — In  technical 
problems  relating  to  heat  engines  the  expansions  and  com- 
pressions are  never  isothermal  or  adiabatic  altho  these 
may  be  the  ideal  conditions  whose  attainment  is  sought. 
Thus  the  pv-curwes  representing  the  actual  changes  of  state 
are  not  represented  by  the  equations  pv=piVx  or  pv^  =  p\V]^- 
It  has  been  shown  that  the  expansion  and  compression 
curves  as  drawn  by  an  indicator  may  often  be  represented 
by  the  equation  pv^'^pivi^,  where  n  is  a  constant.  Under 
these  conditions  the  actual  changes  occurring  in  the  engine 
cylinder  are  polytropic.  Moreover  the  values  of  n  are 
found  to  lie  between  i  and  k. 

Three  methods  for  finding  the  n  of  any  polytropic  traced 
by  an  indicator  will  be  described. 

(i)  Take  any  two  points  on  the  curve  and  measure 
their  p  and  v  coordinates  (always  from  the  axes  of  absolute 
pressure   and   of   absolute   volume).      Then   assuming   the 


CHANGES    OF    STATE    OF    IDEAL   GASES 


57 


curve  to  be  a  polytropic  n  may  be  found  by  substituting 
in  the  formula 

^  log  pi  —  log  p2 
log  F2  — log  Vi 

Exercise  66.  Develop  the  above  formula. 

(2)  The  areas  ABMN  and  ABPQ  (Fig.  5)  may  be  found 
by  means  of  a  planimeter.  Now  if  the  curve  AB  is  a  poly- 
tropic whose  equation  is  pv^  =  piVi^,  then 


n= 


area  ABMN 
area  ABPQ ' 


Exercise  67.  Demonstrate  the  correctness  of  the  above 
described  method  of  finding  //. 

(3)  The  third  and  best  method  does  not  assume  the 
curve  AB  (Fig.  5)  to  be  a  polytropic  but  shows  whether 
it  is  or  is  not  a  polytropic.  If  it  is  not  a  polytropic  this 
method  shows  how  closely  it  approaches  a  polytropic  and 
gives  the  value  of  n  for  the  actual  polytropic  which  best 
approximates  the  given  curve.  It  often  happens  that  a 
given  curve  may  be  very  closely  represented  not  by  one 
but  by   two    separate    polytropics  each  for  its  own   part 


58  THERMODYNAMICS 

of  the  expansion  or  compression.    This  method  of  finding 
n  will  not  only  give  the  value  of  n  for  each  polytropic  but 
will  also  show  the  region  in  which  each  is  applicable. 
A  curve  whose  equation  is 

pv^  =  p\V\^=C,  a  constant 

becomes  a  straight  line  when  plotted  on  axes  of  logio  p 
and  logio  v  instead  of  p  and  v.  For  the  equation  may 
be  written 

(logio  p)=-n{\ogio  z;)+logio  C, 

which  is  the  equation  of  a  straight  line  when  the  variables 
are  (logio/')  and  (logic  t;). 

Thus  if  we  replot  the  curve  AB  (Fig.  5)  upon  log-log 
paper  (i.e.,  paper  ruled  according  to  the  logarithmic  scale 
instead  of  the  usual  decimal  scale)  the  resulting  curve  will 
be  a  straight  line  if  ^S  is  a  polytropic.  In  any  case  the 
best  representative  straight  line  thru  the  plotted  points 
represents  the  log-log  plot  of  a  polytropic  approximating 
the  curve  AB.  The  slope  of  this  straight  line  with  its 
sign  changed  gives  the'value  of  n  in  the  equation  pv"  =  pivi^ 
which  approaches  most  closely  to  the  given  curve  AB. 

Exercise  68.  Prove  the  last  statement. 

The  External  Work  Performed  during  polytropic  changes 
of  state  may  be  computed  either  by  means  of  the  pv-cxirve 
or  by  means  of  the  fundamental  energy  equation. 

Exercise  69.  Show  by  means  of  j  pdV  that  the  external 
work  performed  during  a  polytropic  change  of  state  from  the 
state  pi,  Vi,  Ti  to  the  state  p,  V,  T  is 


CHANGES  OF  STATE  OF  IDEAL  GASES         59 

Exercise  70.  Starting  with  the  equation  aQ=aK+aW, 
show  that  the  external  work  performed  during  a  polytropic 
change  of  state  is 

J{AW)  =  mJic-cXT-T.)  =  ^(T,-T)  =  '^.-^ 

n—i  n  —  i\       li 

Exercise  71.  Show  that 
p,V,-pV 


n  — I 


=  mJ(c-c.)(T-TO 


and  that  the  last  expression  always  gives  a  positive  result  for 
expansion  and  always  gives  a  negative  result  for  compression. 

The  Heat  Supplied  during  a  polytropic  change  of  state. 

Exercise  72.  Show  that  the  heat  supplied  to  an  ideal  gas 
during  a  polytropic  change  of  state  from  pi,  Vi,  Ti  to  p,  V,  T  is 

AQ  =  mc{T-Ti)^mU-\-jr^—\{T-Ti). 

Exercise  73.  Show  that  the  change  in  internal  energy  during 

a  polytropic  change  is 

ipV-PiV: 
AK  =  mcv{l  —  1 1)  =  —  -         " 


/       k-i      ' 


when  expressed  in  B.t.u.  (see  also  Exercise  86)  and  that  therefore 

k  —  i 
whence 

AW:  AK  :AQ  =  k-i  :  i-n  :  k-n. 

Exercise  74.  Show  that  the  expression  for  aQ  in  Exercise  73 
may  be  negative  during  certain  polytropic  expansions. 

Exercise  75.  Compute  by  means  of  E.xercise  69  the  external 
work  done  during  an  adiabatic  change 


6o  THERMODYNAMICS 

(a)  from  lo  pounds  per  square  inch  absolute  and  17.16  cubic 
feet  to  200  pounds  per  square  inch  absolute,  and 

(b)  from  200  pounds  per  square  inch  absolute  and  2  cubic  feet 
to  10  pounds  per  square  inch  absolute. 

Interpret  the  signs  of  the  results. 

Exercise  76.  Air  expands  polytropically  from  300  pounds 
per  square  inch  absolute  and  0.4  cubic  foot  to  15  pounds  per 
square  inch  absolute  with  «  =  3. 

Compute  (o)  the  external  work  done,  (5)  the  heat  supphed, 
(c)  the  specific  heat,  the  mass  times  the  change  in  temperature 
from  the  result  of  (b),  and  thus  the  change  in  internal  energy. 

Exercise  77.  Compute  the  external  work  done  during  an  iso- 
thermal change  of  state  from  20  pounds  per  square  inch  abso- 
lute and  25  cubic  feet  to  100  pounds  per  square  inch  absolute 
and.  5  cubic  feet. 


CHAPTER   III 
GR.\PHICS   OF  THE   pv-FLANE 

Section  X 
PLOTTING   POLYTROPICS 

Plotting  the  Polytropic  Curve. — To  construct  the  cur/e 
pv^  —  piVi",  assume  any  angle  a,  Fig.  6,  and  by  means  of 

(i+tan/3)  =  (i  +  tana)'' 


Fig.  6. 


compute    the    corresponding   angle   /S,    using    the    assigned 
value  of  n.     Set  off  these  angles  as  shown  in  Fig.  6  and  start 

6i 


62  THERMODYNAMICS 

the  construction  by  drawing  pi  and  ^i  from  some  known 
point  on  the  required  polytropic. 

To  demonstrate  the  correctness  of  this  construction  note 
that 

V—Vi 


Vl 


•=tan  a 


Pi  —  P 
and  that =  tan  j(3. 

P 
Hence  v=Vi{i-\-tana) 


and 


Pi 


i+tan/3' 


«i(i  +  tan  a)** 
so  that  K=?i^i"4+I^- 

But  (i+tan/3)  =  (i+tana)'', 

therefore  pv^=p\V\^. 

This  constructi®n  must  be  performed  with  great  accuracy, 
for  in  it  errors  are  cumulative. 

Exercise  78.  Starting  with  (i+tan /3)  =  (i+tan «)"  deduce 
from  it  and  the  construction  illustrated  in  Fig.  6  the  fact  that 

Exercise  79.  Show  that  the  equation  of  the  dotted  curve 
in  Fig.  6  is  ^i'~"  =  ^ii'i~". 

Exercise  80.  How  could  the  points  on  the  curve  found 
by  means  of  the  construction  illustrated  in  Fig.  6  be  caused 
to  lie  closer  together? 


GRAPHICS    OF    THE   />l'-PLANE 


63 


Exercise  81.  Construct  thru  (/>i  =  ioo,  't'i  =  3) 

(a)  an  isothermal  (see  pages  35  and  63). 

(b)  an  adiabatic 

(c)  a  polj'tropic  for  11=1.2 
for  air,  all  on  the  same  diagram. 

Exercise  82.  Plot  a  polytropic  curve  thru  (/>i  =  ioo.  t'i  =  3) 
or  n=i.2  by  direct  substitution  in  the  equation  pv^=piVi". 
Assume  values  of  v  and  compute  p  by  means  of 

logp=  —  n  log  v  +  i\og  pi +11  log  Vi). 

Tabulate  your  computations  as  follows: 


log  V     I   n  log  V 


logp 


A  simpler  and  more  rapid  way  of  plotting  a  polytropic 
involves  the  use  of  log-log  paper.  On  paper  ruled  both 
vertically  and  horizontally  on  the  logarithmic  scale  locate 
the  point  (pi,  vi),  the  given  point  thru  which  the  poly- 
tropic should  pass.  Thru  this  point  draw  a  straight 
line  havnng  a  slope  —n,  where  n  is  the  exponent  of  the 
required  polytropic.  This  line  represents  the  polytropic 
on  the  log-log  plane,  see  page  58.  Now  transfer  as  many 
points  as  may  be  desired  from  this  plane  to  a  plane  ruled 
both  vertically  and  horizontally  on  the  decimal  scale. 
These  points  will  lie  on  the  required  polytropic  in  the 
pv-p\a.ne. 

Construction  of  the  Isothermal  Curve. — The  isothermal 
is  readily  constructed  by  the  method  given  on  page  35. 
Another  method  which  does   not   involve   the  drawing  of 


64 


THERMODYNAMICS 


parallel   lines   is    illustrated   in    Fig.    7.      Here    OA=MN, 
OB  =  PQ,  etc. 


B  M 

Fig.  7. 

Exercise  83.  Show  that  the  construction  illustrated  in  Fig. 
7  yields  a  curve  whose  equation  is  pv  =  piVi. 


Section  XI 
GRAPHICAL  REPRESENTATION   OF  AW,   AK,   AND   AQ 

Graphical    Representation    of     External    Work. — The 

external  work  performed  during  any  change  of  state  equals 

I     pdv.    Therefore  the  area,  Fig.  8, 

bounded  by  the  curve  representing 
the  change  of  state,  the  end  ordinates, 
and  the  z)-axis  represents  the  external 
work  performed. 

In  order  to  determine  the  sign  of 
this    area     the    point     tracing     the 
Fig.  8.  boundary  of  the  area   (the  stylus  of  a 

planimeter)  should   always    start  at 
the  initial  point    of    the    line  representing  the  change  of 


GRAPHICS    OF    THE   pV-PLASE  6$ 

state  (A,  Fig.  8)  and  then  follow  this  line  to  the  final 
point  {p,  v)  and  so  on  around  the  area.  If  the  area  is 
thus  traced  in  the  clockwise  direction  the  area  is  to  be 
considered  positive,  if  in  the  reverse  direction,  negative. 
That  this  rule  of  signs  agrees  with  the  rules  already- 
established  can  be  seen  from  Fig.  8.  Here  an  expansion 
occurs  from  .4  to  B  and  the  area  is  traced  in  the  clockwise 
direction  and  is  thus  positive. 

Exercise  84.  Under  what  conditions  would  the  area  bounded 
by  the  curve  representing  the  change  of  state,  the  abscissas 
of  the  end  points,  and  the  portion  of  the  p-a.xis  intercepted 
between  these  abscissas  represent  the  work  done  during  the 
change  of  state  considered? 

Graphical  Representation  of  Internal  Energy. — The  in- 
ternal energy  of  an  ideal  gas  is  a  function  of  the  tem- 
perature only.  When  an  ideal  gas  has  a  temperature  of 
absolute  zero  it  possesses  no  internal  energy.  An  ideal 
gas  expanding  adiabatically  does  work.  This  work  is 
done  wholly  at  the  expense  of  its  internal  energy.  The 
work  done  thus  equals  the  loss  of  internal  energy.  If 
the  gas  is  assumed  to  expand  adiabatically  from  any  initial 
state  until  its  volume  is  infinite  and  its  pressure  and  tem- 
perature both  zero  the  work  done  will  equal  the  internal 
energy  which  the  gas  possessed  at  the  initial  state  con- 
sidered. 

Graphically,  the  area  under  the  adiabatic  extending  to 
infinity  (Fig.  9)  represents  Ki.  This  internal  energy  must 
always  be  positive  so  that  in  tracing  the  area  we  must 
start  at  the  point  considered  (pi,  vi)  and  follow  the  adia- 
batic and  so  on  around  the  area. 


66 


THERMODYNAMICS 


Exercise  85.  Show  that   the  internal   energy  of  one  pound 
of  an  ideal  gas  whose  state  is  (pi,  Vi,  Ti)  is 


p\Vl 

k-i 


,  a  finite  quantity. 


Graphical  Representation  of  the  Change  in  Internal 
Energy. — The  change  in  internal  energy  due  to  a  change 
of  state  from  px,  vi,  T\  to  p,  v,  T  (Fig.  10)  would  of  course 
equal  the  difference  between  the  areas  K  and  Ki,  or  K—K\. 

Exercise  86.  Show  that  AK  =  K-Ki  =  Jmcv{T—Ti). 


P 


I 
W/////////////////////A^/A 

V 


Fig.  9. 


P 


ipv) 


y//////////M 


Fig.  10. 


As  the  areas  shown  in  Fig.  10  extend  to  infinity  they  cannot 
be  used  to  find  AX"  by  means  of  a  planimeter.  To  rep- 
resent t^K  by  an  area  finite  in  all  its  dimensions  proceed 
as  follows. 

Draw  an  isothermal  thru  the  initial  point  (^1,  'Qi) 
and  an  adiabatic  thru  the  final  point,  Fig.  11.  These 
curves  intersect  in  some  point  C.  Note  that  the  adia- 
batic approaches  the  D-a.\is  more  rapidly  than  the  iso- 
thermal. The  area  under  the  adiabatic  between  the  final 
point   of   the   change   of   state   considered   and   the   inter- 


GRAPHICS    OF   THE    />Z'-PLANE 


67 


section  of  the  adiabatic  and  the  isothermal  equals  AK 
for  the  change  of  state  from  (pi,  vi,  Ti)  to  {p,  v,  T). 

To  prove  that  this  is  so  consider  the  cycle  ABC  A  thru 
which  the  gas  may  be  conceived  to  pass.  It  should  be 
rem.embered  that  the  gas  actually  changes  its  state  only  in 
accordance  with  the  law  represented  by  the  line  AB.  The 
h>^othetical  cycle  ABC  A  returns  the  gas  to  its  initial 
state  at  A. 

In    passing    thru    the    change    from    C  to  ^    the    tem- 


P 


I       AK   %■-- 
V///////////////////A 


Fig.  II. 


perature  (and  therefore  the  internal  energy)  does  not  change. 
Thus  any  gain  in  internal  energy  during  the  change  from 
.1  to  B  must  be  exactly  neutralized  by  an  equal  loss  of 
internal  energy  during  the  change  from  B  to  C.  But  the 
loss  of  internal  energy-  from  J5  to  C  is  represented  by  the 
area  Aiv.  Therefore  this  area  AiiC  also  represents  the 
gain  in  internal  energy'  during  the  change  from  A  to  B. 

This  area  representing  a  gain  must  be  positive  so  that 
we  must  start  tracing  the  area  at  (/>,  v),  then  follow  along 
the  adiabatic  to  C  and  so  on  around  the  area.     An  area 


THERMODYNAMICS 


about  which  the  tracing  point  would  move  in  a  counter- 
clockwise direction  when  following  the  above  rule  would 


Fig.  12, 


be  negative  and   would  indicate  a  loss  of  internal  energy 
during  the  change  of  state  from  {p\,  vi)  to  {p,  v). 


Fig.  13. 


Exercise  87.  Show  that  the  construction  illustrated  in  Fig. 
12  gives  Aii:  for  the  change  of  state  from  {pi,  ih)  to  {p,v). 


GRAPHICS    OF    THE    pV-PLANB 


69 


Graphical  Representation  of  the  Heat  SuppHed. — The 

heat  supplied  during  any  change  of  state  may  be  repre- 
sented by  the  area  included  between  the  line  representing 
the  change  of  state  and  the  adiabatics  thru  the  points 
representing  the  initial  and  the  final  states  of  the  gas,  both 
adiabatics  extending  to  infinity. 

In  Fig.  13  areas  A-\-B    represent   AW,  C-\-D  represent 
K,  B-\-D  represent  Ki. 


(^.^^>   _^B 


I  AQ       - 

m////////////////////////////A 


Fig.  14. 


As  A()=Ai^+APF,  NQ  would  be  represented  by 

\{C^D)-{B^'D)\^\A-^B\     or    A^C. 

Another  area,  finite  in  all  of  its  dimensions,  which  rep- 
resents A(2  is  shown  in  Fig.  14.  The  sign  of  this  area  is 
determined  by  starting  at  the  initial  point  then  following 
the  line  representing  the  change  of  state,  the  adiabatic 
thru  the  final  point,  and  so  on  around  the  area  marked 
AQ.  An  area  traced  in  a  clockwise  direction  is  plus  as 
before. 


70  THERMODYNAMICS 

Exercise  88.  Show  that  the  area  shown  in  Fig.  14  repre- 
sents the  heat  supplied  during  the  change  of  state  represented 
by  AB. 

Exercise  89.  Indicate  areas  on  the  pv-p\a.ne  finite  in  all 
their  dimensions  representing  the  external  work,  the  change  in 
internal  energy,  and  the  heat  supplied  during 

(a)  an  isopiestic  expansion, 

(h)  an  isopiestic  compression, 

(c)  an  isometric  decrease  in  pressure, 

(d)  an  isothermal  compression, 

(e)  an  adiabatic  expansion. 

Note  the  signs  of  these  areas.  Do  they  agree  with  the  signs 
given  in  the  table  on  page  56? 


CHAPTER   IV 
COMPRESSORS 


Section  XII 
SINGLE-STAGE   COMPRESSION 

Piston  compressors  take  in  a  charge  of  gas  at  a  low 
pressure  and  after  compression  to  a  higher  pressure  deliver 
this  gas  into  a  receiver  where  it  is  stored  at  the  higher 
pressure. 

For  an  ideal  compressor  having  no  clearance  these 
processes  may  be  represented  on  the  ^TJ-plane  as  shown 


Fig.  is. 

in  Fig.  15.  The  line  i  2  does  not  represent  an  isopiestic 
change  of  state.  During  this  process  the  mass  of  gas 
considered  does  not  remain  constant.  This  line  repre- 
sents the  gradual  filling  of  the  cylinder  as  the  piston  moves 
toward   the   right.      Thus    during    this    process   the    tem- 

71 


72  THERMODYNAMICS 

perature  and  the  pressure  of  the  gas  remain  constant  but 
the  volume  and  the  mass  of  the  gas  in  the  cylinder  con- 
tinually increase  as  the  piston  moves  toward  the  right. 

When  the  cylinder  is  full  the  inlet  Vtilve  closes  and 
the  piston  moves  toward  the  left,  compressing  the  charge. 
During  this  process  the  mass  of  gas  in  the  cylinder  remains 
constant  (assuming  no  leakage)  and  the  change  of  state 
represented  by  the  line  2  3  may  be  isothermal,  adiabatic, 
or  in  general  polytropic. 

The  compressed  gas  must  now  be  expelled  from  the 
cylinder  by  a  further  motion  of  the  piston  to  the  left.  The 
outlet  valve  opens  and  the  gas  is  displaced  as  shown  by 
the  line  3  4  which  does  not  represent  an  isopiestic  change 
of  state.  During  this  process  the  temperature  of  the 
compressed  gas  is  assumed  to  remain  constant  and  equal 
to  the  temperature  after  compression  at  3. 

The  work  required  to  compress  and  deliver  the  gas  may 
be  computed  irom  Fig.  15.  The  work  done  by  and  against 
the  atmospheric  pressure  on  the  right-hand  side  of  the 
piston  during  the  forward  and  the  backward  stroke  of  the 
piston  are  equal  and  thus  neutralize  each  other.  On  the 
left-hand  side  of  the  piston  the  gas  entering  the  cylinder 
exerts  a  pressure  pi  (or  P2)  upon  the  piston  and  thus  does 
work  represented  by  the  area  under  the  line  i  2.  During 
the  return  stroke  the  gas  exerts  at  first  a  variable  pressure 
(represented  by  the  ordinates  of  the  line  23)  upon  the 
piston,  and  work  represented  by  the  area  under  the  line 
2  3  must  be  done  by  the  machine  upon  the  gas.  Finally 
to  deliver  the  gas  the  work  done  by  the  machine  equals 
the  area  under  the  line  3  4. 

The  total  work  required  for  compression  and  delivery 


COMPRESSORS  73 

is  thus  represented  by  the  area  i  2341.  This  work  must 
be  done  upon  the  gas  and  thus  represents  energy  which 
must  be  supplied  from  some  external  source. 

In  developing  formulas  for  the  computation  of  this  work 
care  must  be  taken  to  note  the  signs  of  the  expressions 
for  the  various  areas  concerned.  Thus  the  area  under 
the  line  2  3  is 

P2V2  loge  — , 

V2 

if  we  assume  isothermal  compression.  As  vz  is  less  than 
V2  this  expression  is  intrinsically  negative.  In  passing 
from  3  to  4  the  work  done  is 

along  the  line  i  2  we  have 

Adding  these  expressions  we  find  as  the  net  work 

1)3  V3 

P2V2+  i—p3V3)  +  p2V2  log€  —=p2V2  loge  — , 

a  negative  quantity,  as  it  should  be,  for  work  is  done  on 
the  gas. 

Another  method  would  be  to  write  all  areas  in  an  in- 
trinsically positive  form,  thus 

1W2  =  P2V2 

Tir  1  ^2 

3W2  =  P2V2  log  — 
'iWs  =  p:iV3 


74  THERMODYNAMICS 

and  then  add  and  subtract  these  expressions  as  indicated 
in  Fig.  15,  thus 

P3'V3-\-p2V2  log P2V2. 

This  gives  a  positive  answer  which  must  of  course  be 
understood  to  represent  the  work  necessary  for  compression 
and  delivery  and  which  must  be  supplied  from  outside 
sources  of  energy. 

Exercise  go.  Show  that  the  work  required  to  compress  and 
deliver  an  ideal  gas  is 

P2V2  lege  — ,  for  isothermal  compression, 
pi 

k  .  .        . 

■ {p?.Vz  —  p-iV2),  for  adiabatic  compression, 

k—i 


P2V2 

n  —  i 


n-\ 

p^ 


,  for  poly  tropic  compression, 


where  p2  and  V2,  and  ps  and  F3  are  the  pressures  and  the  vol- 
umes of  the  gas  at  the  beginning  and  at  the  end  of  the  com- 
pression, respectively,  as  indicated  in  Fig.  15. 

If  no  heat  is  removed  from  the  gas  during  compression 
the  process  represented  by  2  3,  Fig.  15,  would  be  adiabatic. 
If  it  were  possible  to  remove  sufficient  heat  so  that  the 
temperature  of  the  gas  during  compression  would  remain 
constant  and  equal  to  the  temperature  of  the  gas  when 
compression  starts  then  the  process  2  3  would  be  isothermal. 

As  an  adiabatic  is  a  steeper  curve  than  an  isothermal 
more  work  must  be  done  during  adiabatic  than  during 
isothermal  compression.      Therefore  in  practice  the  cooling 


COMPRESSORS  75 

of  the  gas  during  compression  is  important.  This  cooling 
is  effected  by  jacketing  the  cylinder  or  by  injecting  cold 
water  into  the  cylinder.  It  is  however  impossible  to 
remove  sufficient  heat  during  fairly  rapid  compression  to 
effect  even  isothermal  compression  much  less  improve 
on  this  condition.  Practically  the  compression  curve  is 
approximately  polytropic  with  an  exponent  varying,  with 
the  design  and  operation  of  the  compressor,  from  i  to  1.4, 
with  an  average  value  of  1.25. 

Exercise  91.  A  compressor  is  to  take  100,000  cubic  feet 
of  air  per  hour  at  15  pounds  per  square  inch  absolute  and  70^ 
F,  and  deliver  it  at  60  pounds  per  square  inch  absolute.  As 
a  first  approximation  compute  the  horse-power  required  to 
drive  this  compressor  assuming 

(a)  adiabatic  compression 

(b)  polytropic  compression,  n—1.2. 

Exercise  92.  Compute  the  heat  to  be  removed  from  a  gas 
during  compression  in  order  to  produce  a  polytropic  compression 
having  an  exponent  71.  Express  the  result  in  terms  of  the  pres- 
sures and  the  volumes  at  the  beginning  and  the  end  of  com- 
pression and  the  gas  constants.     Solve  by  means  of 

(a)  the  equation  aQ  =  aK+aW, 

(b)  the  specific  heat  of  the  gas. 

Exercise  93.  Assuming  that  the  cooling  water  suffers  an 
increase  in  temperature  of  20°  F  during  its  passage  thru 
the  compressor  described  in  Exercise  91  and  that  the  cooling 
produced  occurs  only  during  compression,  how  much  cooling 
water  must  be  supphed? 

What  expenditure  of  power  does  this  cooling  water  save? 

The  Effect  of  Clearance. — In  all  practical  compressors 
the  piston  cannot  come  into  contact  with  the  cylinder- 
head  at  the  end  of  its  stroke.     The  volume  between  the 


76 


THERMODYNAMICS 


valves  and  the  piston  when  in  its  extreme  position  is  called 
the  clearance  volume.  The  clearance  is  usually  expressed 
as  a  percentage  of  the  displacement  volume  of  the  piston 
which  is  the  volume  swept  thru  by  the  piston. 

The  effect  of  this  clearance  volume  is  shown  on  the 
^zj-diagram  in  Fig.  i6.  After  compression  at  3  the  volume 
of  compressed  gas  in  the  cylinder  is  a  3.  The  piston  can 
expel  only  the  volume  3  4  from  the  cylinder  for  at  4  it 
reaches  the  end  of  its  stroke.  The  volume  a  4  of  com- 
pressed gas  thus  remains  in  the  cylinder  when  the  piston 


A    -I- Displacement  volume- 

^  Clearance  volume 

Fig.  16. 


starts  its  return  stroke  and  expands  during  this  return 
stroke  of  the  piston  as  shown  by  the  process  4  i.  The 
inlet  valves  do  not  open  until  the  point  i  is  reached  at  which 
the  pressure  in  the  cylinder  drops  to  the  intake  pressure. 

The  volume  of  air  admitted  to  the  cylinder  per  stroke 
is  thus  F2— Fi.  This  volume  is  called  the  low-pressure 
capacity  of  the  cylinder. 

Exercise  94.  The  piston  of  a  compressor  has  a  displacement 
volume  of  6  cubic  feet.  The  clearance  is  4  per  cent.  The 
compressor  operates  between  15  and  200  pounds  per  square 
inch  absolute.  Assuming  n=i.2,  what  volume  of  air  does 
the  compressor  take  in  per  stroke? 


COMPRESSORS  77 

Exercise  95.  Show  that  if 

where  pi  is  the  delivery  pressure, 
pi  is  the  intake  pressure, 
V  is  the  displacement  volume, 
Fo  is  the  clearance  volume, 
the  compressor  delivers  no  gas. 

What  power  is  required  to  run  the  compressor  under  these 
conditions? 

Exercise  96.  At  what  pressure  will  delivery  cease  with  iso- 
thermal compression  if  the  intake  pressure  is  14  pounds  per 
square  inch  absolute  and  the  clearance  is  5  per  cent? 

Exercise  97.  Compute  the  volumetric  efBciency  of  a  com- 
pressor (i.e.,  the  ratio  of  the  low-pressure  capacity  to  the  dis- 
placement volume)  in  terms  of  the  clearance  c,  the  initial  and 
the  final  pressures  pi  and  pi,  and  n. 

Exercise  98.  Compute  the  i.h.p.  of  a  double-acting  com- 
pressor with  clearance  in  terms  of  its  low-pressure  capacity, 
the  initial  and  the  final  pressures,  and  n. 

Section  XTII 
COMPOUND    COMPRESSION 

As  some  of  the  above  exercises  show,  the  capacity  of  a 
cylinder  is  diminished  by  the  necessary  clearance.  If  the 
range  of  pressure  is  large  this  capacity  may  be  very  much 
diminished. 

On  account  of  this  decrease  in  volumetric  efSciency  as 
well  as  on  account  of  the  saving  of  work  that  can  be  eflfected, 
compression  to  more  than  six  times  the  intake  pressure 
is  never  attempted  in  a  single  cylinder. 


78 


THERMODYNAMICS 


In  two-stage  compression  the  gas  is  thoroly  cooled 
during  its  passage  from  the  first  cylinder  to  the  second 
one.  The  ideal  conditions  (clearance  neglected)  are  shown 
in  Fig.  17.  Thru  i  are  drawn  an  adiabatic,  a  polytropic, 
and  an  isothermal.  The  area  representing  the  saving  of 
one-stage  isothermal  over  one-stage  adiabatic  compres- 
sion between  the  pressures  pi  and  p2  is  at  once 
evident. 

Suppose  now  that  the  compression  follows  the  polytropic 
law  and  that    the  compression  in  the  first  cylinder  stops 


Fig.  17. 


at  A  (pressure  p').  As  it  is  impossible  to  cool  the  gas 
thoroly  in  the  cylinder  let  the  gas  be  expelled  from 
this  cylinder  (along  AC)  thoroly  cooled  in  an  mter- 
cooler  and  then  returned  to  the  second  cylinder  (along 
CB).  This  diminishes  the  volume  of  the  gas  from  AC 
to  BC  but  its  mass  is  unchanged.  Moreover  by  thoro 
cooling  is  to  be  understood  that  the  gas  has  been  cooled 


COMPRESSORS  79 

to  its  original  temperature  (the  temperature  at  i).  Thus 
B  lies  on  the  isothermal  thru  i  and 

In  the  second  cylinder  the  compression  is  continued  from 
B  io  2  and  the  gas  finally  delivered  at  the  pressure  p2. 
Note  carefully  the  area  representing  the  work  saved  by 
two-stage  poly  tropic  over  one-stage  poly  tropic  compression. 

It  is  now  important  to  determine  the  intermediate  pres- 
sure p',  which  will  yield  the  greatest  saving  of  work. 

The  work  done  on  the  gas  in  the  first  cylinder  is 

in  the  second  cylinder  it  equals 

if  we  assume  the  same  exponent  for  the  law  of  compression 
in  both  cylinders. 

As  the  second  cylinder  must  handle  all  the  gas  supplied 
by  the  first  cylinder  and  as  the  temperature  at  B  equals 
the  temperature  at  i  due  to  perfect  intercooling  we  have 
by  reason  of  pV^mRT 

PiVi  =  P'Vb. 

Thus  the  total  work  required  to  compress  and  deli\-er 
equals 


8o  THERMODYNAMICS 

Exercise  99.  Show  that  if 

p'=\/p,p., 

a  minimum  of  work  is  necessary  for  two-stage  compression. 

Exercise  100.  Show  that  W1  —  W2  when  p'  is  adjusted  for 
minimum  work. 

Exercise  ioi.  When  p'=^pip2  show  that  the  temperature 
of  the  gas  at  the  end  of  the  first  stage  equals  the  temperature 
of  the  gas  at  the  end  of  the  second  stage  of  compression. 

Exercise  102.  Show  that  the  results  obtained  in  Exercises 
99  and  100  are  not  affected  by  whatever  clearance  may  exist 
in  either  or  both  cylinders  of  a  two-stage  compressor. 

Exercise  103.  Compute  the  dimensions  of  the  cylinders  and 
the  power  required  to  drive  a  double-acting  two-stage  air  com- 
pressor designed  to  deliver  3000  cubic  feet  of  air  per  minute 
(measured  at  15  pounds  per  square  inch  absolute)  at  100  pounds 
per  square  inch  absolute.  Assume  the  intake  pressure  to  be 
15  pounds  per  square  inch  absolute,  the  revolutions  80  per 
minute,  the  piston  speed  600  feet  per  minute,  the  clearance 
4  per  cent  in  each  cylinder,  and  the  exponents  of  the  polytropic 
curves  1.3. 

Three-stage  Compression. — When  compression  is  to  be 
carried  to  very  high  pressures  three  stages  may  advan- 
tageously be  employed.  In  order  to  determine  the  two 
intermediate  pressures  which  will  yield  minimum  work  when 
three  cylinders  are  used  the  calculus  may  be  used,  but 
this  method  leads  to  the  investigation  of  a  function  of 
two  independent  variables. 

Instead  we  niay  regard  the  first  two  cylinders  as  con- 
stituting a  two-stage  compressor  and  the  intermediate 
and  high-pressure  cylinders  as  constituting  another  two- 
stage  compressor. 


COMPRESSORS  8l 

Then  if  pi  is  the  initial  pressure, 
p2  the  final  pressure, 

p'  the  pressure  between  the  first  and  the  second, 
and  p"  the    pressure    between    the    second    and    the 
third  cylinders  we  have  for  minimum  work 

p'^Vpip"     and     />"=\/7>2. 
Solving  these  equations  for  p'  and  p"  we  find 


P'=^prp2 
and  p"=^pip2^. 

Exercise  104.  Find  these  values  of  p'  and  p"  by  means  of 
the  calculus. 

Turbo-compressors. — In  turbo-compressors  the  gas  is 
compressed  by  the  action  of  rotating  blades.  The  .gas 
is  not  confined  in  cylinders  by  means  of  pistons.  Under 
these  conditions  the  indicator  cannot  be  used  to  find  the 
indicated  horse-power. 

The  work  required  to  compress  and  deliver  a  gas  by  means 
of  a  turbo-compressor  can  be  computed  from  the  follow- 
ing experimental  data. 

(i)  The  weight  and  the  temperature  rise  of  the  cooling 
water  used, 

(2)  the  temperature  rise  of  the  compressed  gas, 

(3)  the  weight  of  the  compressed  gas. 

Let  pi,  Vi,  Ti,  represent  the  initial  state,  and 
P2,  V2,  T2,  the  final  state  of  the  gas, 
Wc,  the  work  of  compression, 
W,  the  work  required  to  compress  and  deliver 
the  gas,  both  in  foot-pounds, 


82  THERMODYNAMICS 

Q,  the  heat  extracted  during  compression,  in 
B.t.u. 

Then  as  AQ^AK+AW 

we  have  —Q  =  mCt,{T2—Ti)  —  —Wc, 

or  Wc=JQ+Jmc,(T2-Ti). 

No  matter  what  the  law  of  compression  may  be 
W=Wc-\-p2V2-piVi, 
thus  W  =  JQ-\-JmUT2-Ti)+p2V2-piVi. 

But  pV=mRT 

so  that       W  =  JQ-\-Jmc,(T2-Ti)+mR{T2-Ti) 
or  W=JQ-\-mJcj,{T2-Ti). 

Exercise  105.  A  turbo-compressor  delivers  per  minute  at 
a  temperature  of  170°  F,  16,000  cubic  feet  of  air,  measured 
at  14.7  pounds  per  square  inch  absolute.  Assuming  that  no  heat 
is  extracted  during  the  compression  find  the  indicated  horse- 
power of  this  compressor. 


CHAPTER  V 

GAS    CYCLES 

Section  XIV 

INTRODUCTION 

The  state  of  an  ideal  gas  is  determined  by  its  pressure, 
volume,  and  temperature.  The  process  thru  which  a 
gas  passes  when  it  changes  its  state  may  conveniently  be 
represented  by  a  line  on  the  pV -plane.  Thruout  such 
a  process  the  mass  of  gas  considered  must  remain  con- 
stant. The  changes  in  temperature,  not  represented  on 
the  />F-plane,  may  be  followed  by  means  of  the  equation 

pV=mRT. 

A  series  of  processes  which  after  completion  leave  the 
gas  in  its  initial  state  is  called  a  cycle.  Thus  after  the 
gas  has  passed  thru  a  cycle  the  pressure,  the  volume,  the 
temperature,  and  of  course  the  internal  energy  have  all 
returned  to  their  initial  values. 

In  Fig.  1 8,  let  the  lines  12,23,31  represent  the  processes 
thru  which  the  given  mass  of  gas  passes  in  order  to 
complete  the  cycle  i  2  3  i.  The  work  performed  by  the 
gas  during  the  process  i  2  is  represented  by  the  area  A  1  2  B. 
The  work  performed  on  the  gas  during  the  process  2  3  is 
represented  by  C  3  2  J5  and  for  the  process  3  i  by  ^4  i  3  C 

83 


84  THERMODYNAMICS 

The  net  work  obtained  from  the  gas  would  thus  be  rep- 
resented by 

A  1  2  B  —  C  3  2  B  —  A  I  3  C, 

or  as  it  may  also  be  expressed, 

A  I  2  B+B  2  T,C+C  ^  I  A. 

The  net  work  is  thus  the  area  enclosed  between  the 
lines  representing  the  processes  thru  which  the  gas  passes 
during  the  cycle,  the  shaded  area  in  Fig.  i8. 


As  will  be  shown,  a  gas  cannot  pass  thru  a  cycle 
unless  heat  is  supplied  to  it  and  is  rejected  by  it  during 
some  of  the  processes  of  the  cycle.  Moreover  when  work 
is  done  by  the  gas  passing  thru  a  cycle  more  heat  is 
invariably  supplied  to  it  than  is  rejected  by  it  and  the 
difference  between  the  supplied  and  the  rejected  heats 
must  equal  the  heat  equivalent  of  the  work  done  by  the 
gas.  This  is  of  course  a  consequence  of  the  law  of  con- 
servation of  energy. 

A  heat  motor  is  any  machine  by  means  of  which  heat 
energy  may  be    transformed  into  mechanical  energy.     In 


GAS   CYCLES  8$ 

all  known  heat  motors  the  following  three  prime  elements 
appear  in  some  form  or  other: 

(i)  A  hot  body,  the  source  of  heat, 

(2)  A  working  substance, 

(3)  A  cold  body,  the  receiver  of  heat. 

During  the  operation  of  a  heat  motor  heat  energy  leaves 
the  hot  body,  passes  to  the  working  substance  by  means 
of  which  some  of  the  heat  energy  is  transformed  into  mechan- 
ical energy  and  the  remaining  heat  energy  is  rejected  by 
the  working  substance  to  the  cold  body  in  order  that  the 
working  substance  may  return  to  its  initial  state  ready 
to  resume  the  cycle. 

The  working  substance  may  be  a  solid,  a  liquid,  or  a 
gas.  It  is  conceivable  that  a  long  bar  of  iron  may  act  as 
the  working  substance  of  a  heat  motor.  Assume  one  end 
of  the  bar  to  be  firmly  fixed  while  the  other  end  acts  as  a 
pawl  upon  a  ratchet-wheel  with  teeth  of  small  pitch.  The 
work  performed  may  be  stored  as  potential  energy  in  a 
weight  lifted  by  means  of  a  cord  wound  about  a  drum 
fixed  to  the  axle  of  the  ratchet-wheel.  If  heat  is  supplied 
to  the  bar  of  iron  (the  working  substance)  by  means  of  a 
flame  (the  hot  body)  the  bar  expands  and  work  is  done. 
In  order  to  return  the  bar  to  its  initial  state  so  that  the 
cycle  may  be  repeated  the  bar  must  now  be  cooled  and 
heat  must  be  rejected  to  the  cold  body  which  may  be 
the  atmosphere.  Heat  motors  operating  on  this  prin- 
ciple have  been  used  to  wind  freak  clocks  by  means  of  the 
daily  changes  in  the  temperature  of  the  atmosphere. 

In  the  discussion  of  any  cycle  the  following  points  should 
be  considered. 

(i)  The  relations  between  the  states  of  the  gas  at  the 


86  THERMODYNAMICS 

beginning  and  at  the  end  of  the  various  processes  involved 
in  the  cycle. 

(2)  The  total  heat  supplied  to  the  gas  by  the  hot  body 
and  rejected  by  the  gas  to  the  cold  body. 

(3)  The  net  external  work  performed  by  the  gas  during 
the  cycle. 

(4)  The  efficiency  of  the  cycle. 

It  should  be  remembered  that  the  algebraic  sum  of  the 
heats  supplied  to  the  gas  during  the  cycle  (this  to  include 
the  heats  rejected  by  the  gas  which  are  negative)  must 
equal  the  heat  equivalent  of  the  work  performed  by  the  gas. 

The  efficiency  of  a  cycle  is  defined  by  the  ratio 

output  _      heat  converted  into  work 
input      heat  supplied  by  the  hot  body" 

The  more  important  gas  cycles  which  have  been  pro- 
posed or  used  will  be  discussed  in  this  chapter.  For  con- 
venience they  have  been  grouped  as  cycles  of  hot-air  engines 
and  of  internal-combustion  engines. 

Section  XV 

CYCLES    OF    HOT-AIR   ENGINES 

The  Carnot  Cycle. — No  attempt  has  ever  been  made 
to  build  a  heat  motor  operating  on  a  Carnot  cycle.  Its 
mean  effective  pressure  is  very  small  so  that  a  very  large 
engine  would  yield  but  little  power.  Theoretically  this 
cycle  is  however  very  important.  Its  efficiency,  as  will 
be  shown,  cannot  be  surpassed  by  that  of  any  other  cycle. 
It  thus  serves  as  a  standard  by  means  of  which  other 
cycles  should  be  judged. 


GAS   CYCLES 


87 


The  Carnot  cycle  consists  of  two  isothermal  and  of  two 
adiabatic  processes  as  shown  in  Fig.  19.  The  gas  to  be 
forced  thru  this  cycle  should  be  conceived  as  placed 
in  a  heat-insulated  cylinder,  Fig.  19,  closed  at  one  end  by 
a  non-conducting  piston  and  at  the  other  end  by  a  con- 
ducting plate  A.  The  mass  of  gas  remains  in  the  cylinder 
thruout    the    cycle.      The    end  A   of    the    cylinder    may 


//////////////////////////////f"//////////////////// 


■<-P 


Fig.  19. 


be  covered  at  will  by  either  the  hot  body,  E,   the  cold 
body,  C,  or  a  non-conducting  plate,  /. 

Assuming  the  cycle  to  start  with  the  gas  at  its  greatest 
volume  and  at  its  least  pressure  (i,  Fig.  19)  the  gas  must 
first  be  compressed  along  the  isothermal  i  2  at  the  tem- 
perature of  the  cold  body  Tq-  The  work  required  for 
this  compression  must  be  furnished  by  the  machine.  It 
may  be  conceived  to  be  derived  from  energy  stored  as 
kinetic  energv  in  the  flv-wheel  of  the  machine.      Moreover 


88  THERMODYNAMICS 

the  force  exerted  on  the  piston  P  must  be  conceived  as 
continually  changing  so  as  to  remain  only  very  slightly 
in  excess  of  the  force  exerted  on  the  piston  by  the  con- 
fined gas.  Thus  the  motion  of  the  piston  is  very  slow  and 
the  temperature  of  the  gas  rises  but  very  slightly  above 
Tc  before  the  heat  generated  by  the  compression  flows 
to  the  cold  body  C  which  must  be  placed  at  A  during  this 
process. 

At  2  the  law  of  compression  should  change  to  an  adia- 
batic  one.  The  cold  body  C  must  be  removed  from 
A  and  the  non-conducting  plate  /  substituted  for  it. 
The  work  required  for  compression  is  still  derived  from  the 
machine  but  the  temperature  of  the  gas  now  rises  from 
Tc  to  Th,  the  temperature  of  the  hot  body.  When  the 
temperature  Th  is  reached  (at  3)  the  isothermal  expansion 
should  begin. 

Now  the  force  P  exerted  by  the  mechanism  upon  the 
piston  should  remain  always  very  slightly  less  than  the 
force  exerted  by  the  gas.  Thus  P  must  vary.  The  gas 
now  drives  the  piston  and  the  work  done  must  be  partly 
stored  in  the  machine  for  future  use  in  compressing  the 
gas  and  partly  rejected  by  the  machine.  The  isothermal 
expansion  can  occur  only  if  A  is  now  in  contact  with  the 
hot  body  H  so  that  the  heat  transformed  into  work  may 
be  supplied  to  the  gas  by  conduction  as  soon  as  the  tem- 
perature has  fallen  ever  so  slightly  below  Th- 

To  complete  the  cycle  H  is  removed  and  I  substituted 
at  A  when  the  state  4  is  reached.  The  gas  continues  to 
expand  and  do  work  at  the  expense  of  its  internal  energy 
and  its  temperature  drops  to  Tc  when  state  i  is  reached. 

No  engine  reproducing  exactly  the  processes  above  de- 


GAS   CYCLES 


89 


scribed  could  be  built.  The  condition  of  practical  equi- 
librium involved  during  isothermal  transfers  of  heat  make 
this  impossible.  However  engines  approximating  these 
conditions  may  be  conceived.  So  it  is  with  all  heat  engines. 
The  ideal  conditions  sought  are  never  attained.  The  closer 
the  actual  cycle  approaches  the  ideal  the  more  nearly  will 
the  ideal  efficiency  be  realized. 

From  the  above  description  of  the  Carnot  cycle  it  is 
evident  that  certain  relations  must  exist  between  the  pres- 
sures and  between  the  volumes  of  the  gas  at  i,  2,  3,  and 
4  in  order  that  the  processes  involved  may  form  a  closed 
cycle.  For  example,  the  isothermal  expansion  must  cease 
at  just  the  right  point  in  order  that  the  following  adiabatic 
expansion  may  return  the  gas  to  its  initial  state  at  i. 

From  the  process  2  3  we  have 

n     \Th)     \V2 
and  from  the  process  4  i 

Ti     \Tc)     \V4 

therefore  -^zr^=j^. 

V2     ^  1 

Exercise  106.  What  relation  must  exist  between  the  pressure 
at  the  beginning  and  at  the  end  of  the  various  processes  of  a 
Carnot  cycle? 

•  Exercise  107.  Assuming  the  temperature  of  the  source  as 
500°  F  and  of  the  receiver  as  70°  F,  the  displacement  volume 
of  the  piston  as  10  cubic  feet  with  a  "clearance"  volume  of 
2  cubic  feet,  find  the  highest  pressure  reached  in  a  Carnot  cycle 
operated  under  the  above  conditions  provided  the  lowest 
pressure  is  1 5  pound's  per  square  inch  absolute. 


go  THERMODYNAMICS 

The  heat  supplied  during  the  whole  Carnot  cycle  illus- 
trated in  Fig.' 19  consists  of  the  sum  of  the  heats  supplied 
uuring  the  various  processes.     Thus 

I  Fc 

lQ2  =  jplVi  logey^ 

2(?3  =  0 

3Q4  =  jp3V3l0gty- 

so  that  iQi^^piVi  loge  77^4-7^3^3  loge  77-. 

J  V\      J  Vz 

I  V2      .       .        . 

Note    that    —pi  V\  loge  77—    is    intrinsically  negative    for 
J  Vi 

V2<V\.  Thus  iQi  as  written  above  represents  the  dif- 
ference between  the  heat  supplied  by  the  hot  body  and 
the  heat  rejected  to  the  cold  body.  It  is  the  heat  which 
disappears  as  heat  and  reappears  as  mechanical  energy. 

It  is  important  to  simplify  the  expression  for  \Qi  as 
follows : 

l(?l  =  ;7 (  P3  Vz  loge  ^- pi  Vi  loge  yA 
I  V  A 

=  j{pzVz  —  piVi)\ogt—. 

Exercise  108.  Show  that  the  heat  conv'erted  into  work  during 
the  operation  of  a  Carnot  cycle  equals 

—{Th—Tc)  loge/-, 

Vi     Vi    p2     pz 

where  r=r;-=— -  = — = — . 

F2      Vz     pi     pi 


GAS   CYCLES 


91 


Exercise  109.  Compute  from  Fig.  19  the  area  representing 
the  work  dehvered  during  a  Carnot  cycle  and  show  that  the 
result  is  equivalent  to  the  results  expressed  in  Exercise  loS. 

Exercise  no.  Show  that  the  eflaciency  of  a  Carnot  cycle  is 

Th-Tc 


Fig.  20. 


until 


The  Stirling  Cycle. — The  Stirling  cycle  (Fig.  20)  con- 
sists of  two  isothermals  combined  with  two  isometric  proc- 
esses. This  cycle  affords  larger  mean 
effective  pressures  and  thus  more 
power  for  the  same  cylinder  volume 
than  the  Carnot  cycle. 

Under  ideal  conditions  the  gas 
passing  thru  the  processes  of  a 
Stirling  cycle  should  be  compressed 
while  in  thermal  contact  with  the 
cold  body  (process  i  2,  Fig.  20),  then 
heated  while  its  volume  remains  constant,  2 
its  temperature  is  Th,  the  temperature  of  the  hot  body, 
next  expanded  isothermally  while  in  contact  with  the  hot 
body,  process  3  4,  and  finally  cooled  at  constant  volume 
to  the  temperature  of  the  cold  body,  process  4  i. 

In  this  cycle  the  heat  supplied  during  the  process  2  3  is 

2Q3  =  fnCviTii—Tc), 

and  the  heat  supplied  from  4  to  i  is 

4Qi  =  niCtiTc—TH). 

As  the  last  expression  is   intrinsically   negative   {Tc<Th) 
it  represents  heat  rejected.     Also 


92  THERMODYNAMICS 

Stirling  proposed  to  store  the  heat  rejected  from  4  to  i 
and  utilize  this  heat  during  the  process  2  3  in  order  that 
less  heat  need  be  drawn  from  the  hot  body,  thus  improving 
the  efficiency.  Stirling  did  this  by  means  of  a  regenerator. 
The  regenerator  may  be  conceived  as  a  cylinder  filled  with 
metal  gauze.  One  end  of  this  cylinder  remains  at  the 
temperature  of  the  hot  body,  the  other  at  the  temper- 
ature of  the  cold  body.  The  hot  gas  passing  thru  this 
cylinder  from  the  hot  towards  the  cold  end,  heats  the 
gauze  so  that  the  gas  loses  its  heat  gradually  with  gradually 
falling  temperature  and  leaves  the  cylinder  cold.  Reversing 
the  flow,  the  cold  gas  entering  the  cold  end  of  the  cylinder 
would  (at  least  under  ideal  conditions)  pick  up  the  heat 
previously  deposited  in  the  gauze  and  would  leave  the 
cylinder  hot  without  drawing  upon  the  source  of  heat 
(the  hot  body). 

Stirling  built  and  operated  engines  based  upon  the  above- 
described  cycle.  One  of  these  engines  indicating  50  horse- 
power is  said  to  have  realized  a  thermal  efficiency  of  30 
per  cent  with  1.7  pounds  of  coal  per  i.h.p.  hour. 

Exercise  hi.  Show  that  the  work  performed  by  the  gas 
during  a  Stirling  cycle  is 

m{TH-Tc)R\oger, 
where 

Vi    V2    pi    pi 

Exercise  112.  Show  that  with  an  ideal  regenerator  the 
efficiency  of  the  Stirling  cycle  is 

Th-Tc 
Th     ' 


GAS   CYCLES 


93 


Exercise  113.  Compute  the  efficiency  of  the  Stirling  cycle 
without  regeneration. 

The  Ericsson  Cycle. — The  Ericsson  cycle  consists  of  two 
isothermal  and  of  tw'o  isopiestic  processes.  In  Fig.  21 
the  cycle  is  i  2  3  4  i. 

The  mass  of  gas  passing  thru  the  cycle  may  be  con- 
ceived to  remain  in  the  cylinder  during  the  whole  cycle. 
It  is  first  compressed  while  its  temperature  is  maintained 
at  the  temperature  of  the  cold  body,  process  i  2,  Fig.  21. 


Fig.  21. 


The  gas  is  then  heated  while  its  volume  increases  at  con- 
stant pressure  until  its  temperature  is  Th,  the  temperature 
of  the  hot  body,  process  2  3.  Next  isothermal  expansion 
at  a  temperature  Th  follows  during  which  the  gas  must 
be  heated  from  an  external  source,  the  hot  body.  Finally 
the  gas  is  cooled  at  constant  pressure  to  a  temperature 
Tc  the  volume  meanwhile  reducing  from  F4  to  Vi. 

Practically  the  gas  cannot  be  made  to  pass  thru  the 
above-described  processes  in  one  cylinder  for  this  would 
require  that  the  cylinder  be  heated  and  cooled  from   a 


94  THERMODYNAMICS 

temperature  Tc  to  Th,  and  back  to  Tc  during  each  cycle. 
This  would  entail  a  great  waste  of  heat. 

Ericsson  invented,  designed,  and  built  engines  operating 
on  the  above  cycle  in  the  following  manner.  The  air  was 
first  compressed  as  nearly  isothermally  as  possible  in  one 
cylinder  of  the  engine.  The  action  of  this  cylinder  would 
be  shown  on  a  />F-diagram  (Fig.  21)  by  the  lines  b  1,  i  2,  2  a. 
The  compressed  air  is  now  heated  while  on  its  way  to  the 
second  cylinder  of  the  engine,  the  working  cylinder.  The 
compressed  and  heated  air  (equal  in  mass  to  the  air  dis- 
charged by  the  compressing  cylinder)  is  now  admitted  to 
the  working  cylinder  (process  a  3,  Fig.  21)  expanded  while 
heat  is  supplied  to  it  by  the  hot  body,  process  3  4,  and 
then  expelled  from  the  working  cylinder  by  the  piston, 
process  4  h.  After  leaving  the  working  cylinder  while  on 
its  way  to  the  compressing  cylinder  preparatory  to  resuming 
"the  cycle  the  air  is  cooled  at  constant  volume  so  that  the 
mass  of  air  having  a  volume  of  F4  when  it  leaves  the  work- 
ing cylinder  enters  the  compressor  with  a  volume  Vi. 

Exercise  114.  Show  that  the  heat  which  must  be  supplied  to 
the  gas  during  the  process  2  3,  Fig.  21,  equals  the  heat  which 
must  be  withdrawn  during  the  process  4  i,  and  thus  that  a 
regenerator  may  be  used  to  store  the  heat  rejected  along  the 
process  4  i  for  use  during  the  process  2  3. 

Exercise  115.  Compute  the  efficiency  of  the  Ericsson  cycle 
(a)  with,  {b)  without  regeneration. 

The  Joule  Cycle. — All  engines  operating  on  the  cycle  of 
Ericsson  and  of  Stirling  failed  to  give  satisfaction  in  service 
on  account  of  the  failure  of  the  metal  plates,  which  are 
exposed  on  one  side  to  the  hot  gases  of  the  furnace  and  on 
the  other  to  the  air  in  the  cylinder.    The  air  in  the  work- 


GAS   CYCLES 


95 


ing  cylinder  was  also  practically  at  rest  so  that  the  transfer 
of  heat  to  it  was  very  slow.  The  regenerators  were  also 
liable  to  rapid  deterioration. 

Joule  proposed  a  cycle  which  overcame  these  difficulties. 
His  cycle  consisted  of  two  adiabatics  and  two  isopiestics, 
Fig.  22.  The  compression  and  the  expansion  being  adia- 
batic,  no  attempt  is  made  to  heat  or  cool  the  air  during  these 
processes.  Heat  is  supplied  to  the  air  only  during  the 
processes  2  3  and  4  i.     That  these  transfers  of  heat  may 


Fig.  22. 


be  effected  under  the  best  conditions  Joule's  engine  con- 
sisted of  two  cylinders  and  the  heating  and  cooling  occurred 
during  the  passage  of  the  air  from  one  cylinder  to  the  other. 

A  diagrammatic  sketch  of  the  engine  is  shown  in  Fig.  23. 
The  volume  of  the  gas  during  compression  is  always  less 
than  the  volume  of  the  same  mass  of  air  during  expansion 
(see  Fig.  22)  so  that  if  the  pistons  have  the  same  stroke 
the  diameter  of  the  working  cylinder  must  be  greater  than 
the  diameter  of  the  compressing  cylinder. 

Assuming  ideal  conditions  and  cylinders  without  clearance 
the  states  of  the  gas  during  the  performance  of  the  cycle 


96 


THERMODYNAMICS 


and  the  corresponding  positions  of  the  gas  in  the  engine 
would  be  as  follows.  Intake  into  the  compressor  occurs 
along  b  I  (Figs.  22  and  23),  the  temperature  being  Ti;  delivery 
from  the  compressor  occurs  along  2  a,  temperature  T2 
where  T2>Ti  owing  to  the  adiabatic  compression.  The 
gas  is  now  heated  during  its  passage  thru  the  "  hot 
body  "  leaving  it  at  a  temperature  T3  and  passing  into 
the  working  cylinder  a  3,  Tz>T2.  Adiabatic  expansion  in 
the  working  cylinder  cools   the   gas  to  a  temperature  7*4 


Cold  Body 


at  this  temperature  it  is  delivered  along  4  6  to  the  "  cold 
body  "  there  to  be  further  cooled  to  a  temperature  Ti 
{Ta>Ti),  preparatory  to  reentering  the  compressor  with 
a  volume  Vi  =  b  i . 

The  temperature  of  the  cold  body  under  ideal  conditions 
must  be  Ti  so  that  the  gas  coming  in  contact  with  it  at 
a  temperature  T4,  may  leave  at  a  temperature  Ti.  Simi- 
larly the  temperature  of  the  hot  body  must  be  T3,  altho 
the  gas  reaches  it  at  a  temperature  r2(<r3). 

The  difference  between  the  conditions  of  heat  transfer 


GAS   CYCLES  97 

in  this  cycle  and  the  cycles  so  far  considered  should  be 
carefully  noted.  In  the  other  cycles  the  gas  was  always 
at  the  same  temperature  as  either  the  cold  body  or  the 
hot  body  during  the  transfer  of  heat,  a  less  practical  but 
a  theoretically  more  perfect  condition  and  one  which  as 
we  shall  see  leads  to  a  higher  efficiency.  Whenever  a 
transfer  of  heat  from  a  body  at  higher  to  one  of  lower 
temperature  occurs  energy  is  invariably  wasted. 
For  the  analysis  of  the  Joule  cycle  note  that 

pi  —  p4:  =  pc,  the  pressure  in  the  "  cold  body", 

p2  =  p3  =  pH,  the  pressure  in  the  "  hot  body". 

We  may  obtain  a  relation  between  the  volumes  by  means 
of  either  the  processes  i  2  and  3  4  or  i  4  and  23.  In  the 
latter  the  temperatures  are  involved — about  these  we  as.  yet 
know  nothing.  The  former  gives  the  relations  of  the 
volumes  to  the  pressures.     Thus 


P^^(YlY=(Yl 
pH     \Vx)       \F4 


whence 


V2      Vz 


Now  using  the  processes  i  4  and   2  3  and  remembering 
that  in  pV=^mRT,  p,  in,  and  R  are  constant,  we  find 

,  n  T3 

whence  ;=r=;=-. 

i  1     J  2 


qS  thermodynamics 

The  heats  suppHed  durinff  the  four  processes  are 

2Q3  =  mCp{T3—T2) 
3<24  =  0 

4Q1  =  mcp(Ti  —  7^4)  =  —  mCp{T4,—  Ti). 

The  work  performed  by  the  gas  durmg  each  of  the  four 
processes  is 

TT.      p\Vi  —  p'2V2         P2V2  —  P1V1 

\W2  = 7 = 7 

k—i  k—i 

2W3  =  Ph{V3-V2) 

„  p3Vs—p4:V4: 

3W4  =  - 7 

k—I 

4Wi  =  Pc{Vi-V4)=-pc{V4-Vi). 

Exercise  116.  Find  the  net  work  performed  by  the  gas 
during  a  Joule  cycle  by  means  of  two  areas  and  show  that  the 
result  equals  the  sum  of  the  four  areas  given  above. 

Exercise  117.  Find  an  expression  for  the  net  work  performed 
during  a  Joule  cycle  in  terms  of  the  temperatures. 

To  find  the  efficiency  of  the  Joule  cycle  the  result  of 
Exercise  117  is  more  useful  than  the  result  of  Exercise  116. 
In  general  the  efficiencies  of  a  cycle  can  best  be  computed 
from  the  heat  than  from  the  standpoint  of  work  directly. 
Thus  the  efficiency  of  the  Joule  cycle  is 

mcp{T3—  T2)  —  mcp{T4—  Ti) _    _T4—Ti 
mcp{Tz—T2)  T3—T2 


GAS   CYCLES  99 

This  expression  may  be  simplified  by  means  of  the  relation 
between  the  temperatures  already  deduced,  namely, 

Ti     T2 

From  this  relation  by  subtracting  unity  from  each  member 

^    .  n-Ti     TS-T2 

we  obtain  — :^ =  — ^ , 

i  1  I  2 

U-Ti^Ti 
TZ-T2     T2 

Therefore  the  efficiency  reduces  to 

I-  — =1-  — =  i-f^ 
T2  Tz  \pH, 


or 


k 


It  is  important  to  note  that  this  efficiency  is  less  than 
the  efficiency  of  a  Carnot  cycle  operating  between'  the 
same  hot  and  the  same  cold  bodies.  The  efficiency  of 
the  Carnot  cycle  is 


Th-Tc 
T„ 

=  1- 

Tc 
Th' 

or  in 

the  notation  of  this  case 

Ti 

I  — 

T3 

As 

T3>T2, 

Tx 
Ts 

^2 

and 

^3 

Ti 
T2 

Exercise  ii8.  Show  that  Ti>Ti  and  that  therefore  the 
efficiency  of  the  Joule  cycle  is  less  than  the  efficiency  of  the 
Carnot  cycle  operating  between  the  same  extreme  temperatures. 


lOO  THERMODYNAMICS 

Condition  Necessary  for  Greatest  Efficiency. — It  should 
be  remembered  that  when  heat  is  supphed  to  (and  with- 
drawn from)  the  gas  during  isothermal  processes  and  only 
during  such  processes  that  the  efficiencies  of  the  cycles 
(Carnot,  Stirling,  Ericsson,  the  last  two  with  regeneration) 
are  all  the  same  and  all  greater  than  the  efficiency  of  the 
Joule  cycle.  It  will  be  shown  that  no  cycle  can  exceed 
in  efficiency  the  first-mentioned  cycles.  In  these  cycles 
thermal  equilibrium  always  exists  and  this  renders  the 
exact  fulfilment  of  the  conditions  involved  impossible  in 
practice. 

Cycles  Composed  of  Two  Pairs  of  Polytropics. — All 
cycles  so  far  considered  have  been  composed  of  four  poly- 
tropics, of  which  the  opposite  pairs  have  the  same  ex-ponent. 
In  general  we  may  assume  the 

process  i  2  to  be  the  poly  tropic  p\Vi^=  p2V2^=piP', 
process  2  3  to  be  the  poly  tropic  p2V2^  =  psv-i"  =  pv^, 
process  3  4  to  be  the  poly  tropic  pzV'A^=  piVi^=  pv^, 
process  4  i  to  be  the  polytropic  p4^Vi^  =  pivi^  =  pv^. 

To  establish  the  general  relation  between  the  volumes 
place  the  product  of  the  first  members  of  these  equations 
equal  to  the  product  of  the  second  members.     Thus 

Plp2p3p4.  Vl''V'f'Vs''Vi"'=p2p3p4pl  Z'^A'a'^t'Ar, 

or  (viVsT  e^'21'4'")  =  (vov^r  (I'svir, 

whence  (tail's)""'"  =(^2^4)"-'^ 

or  ViV3=V2V4. 


GAS   CYCLES  lOI 

Similarly  the  relation  between  the  pressures   may  be 

obtained  as  follows 

1  1 

1  i. 

p2"*V2  =  pZ*Vz 

1  1 

prv4=pi"'vi, 

1  i  11 

whence  {pipzT{p2pAT  =  (p2p4T(p3pi  )*" 

1_1  i_i 

(pipsr~"'={p2p4r~"' 

or  PlP3=P2P4. 

These  relations  are  the  same  as  those  already  established 
for  each  individual  cycle  discussed  in  this  section.  They 
do  not  depend  upon  the  substance  which  passes  thru 
the  cycle  but  simply  upon  the  laws  of  change  of  state 
followed. 

The  relation  between  the  temperatures  however  depends 

upon  the  nature  of  the  substance  used.    For  ideal  gases 

the  relation  between  the  temperatures  may  be  developed 

as  follows. 

p\V\=RTi     and     psV3  =  RT3 

therefore  (pip3){viV3)  =  R'iTiTs). 

Similarly  {p.2p4){v2Vi)  =  RHT2T4). 

Thus  by  reason  of  the  above-established  relations  between 
the  volumes  and  the  pressures 

TiT3  =  T2T4. 


I02  THERMODYNAMICS 

Section  XVI 

HEAT  PUMPS   OR  REFRIGERATING   MACHINES 

The  cycles  and  engines  considered  in  the  last  section 
were  all  cycles  and  engines  utilized  as  heat  motors,  the 
object  being  the  transformation  of  heat  energy  into  mechan- 
ical energy.  In  all  cases  heat  passed  from  the  hot  body 
to  the  cold  body  and  during  the  transfer  thru  the 
working  substance  some  of  the  heat  leaving  the  hot  body 
was  transformed  into  mechanical  energy  thus  never  reaching 
the  cold  body. 

If  7]  represents  the  efSciency  of  the  cycle  considered  and 
Q  the  heat  units  leaving  the  hot  body  then  rjQ  represents 
the  heat  units  transformed  into  mechanical  energ}^  and 
{i  —  v)Q  the  heat  units  rejected  to  the  cold  body. 

Exercise  119.  An  ideal  engine  operating  under  a  Joule  cycle 
with  air  under  extreme  pressures  of  40  and  100  pounds  per  square 
inch  absolute  will  reject  what  fraction  of  each  B.t.u.  supplied 
by  the  hot  body  to  the  cold  body,  which  is  maintained  at  70°  F? 

If  the  gas  in  any  of  the  cycles  already  considered  could 
be  forced  to  change  its  state  in  such  a  manner  that  the 
point  representing  the  state  on  the  /?F- plane  would  move 
around  the  cycle  in  a  counter-clockwise  direction  then 
instead  of  yielding  mechanical  energy,  mechanical  energy 
would  have  to  be  supplied  to  the  gas.  Also  instead  of 
rejecting  heat  to  the  cold  body  the  gas  would  withdraw 
heat  from  the  cold  body  and  the  hot  body  would  receive 
heat  from  the  gas  instead  of  supplying  it  with  heat. 

Moreover  the  heat  reaching  the  hot  body  would  equal 
the   heat   withdrawn   from   the   cold   body   plus   the   heat 


GAS   CYCLES 


103 


equivalent  of  the  net  work  done  on  the  gas  or  the  work 
required  to  run  the  engine  under  these  conditions. 

If  the  Carnot  cycle  (Fig.  19)  is  traced  in  the  counter- 
clockwise direction  the  temperature  of  the  hot  body  still 
persists  along  the  process  4  3  and  the  temperature  of  the 
cold  body  along  the  process  2  i  is  still  the  temperature  at 
which  heat  is  now  transferred  from  the  cold  body  to  the 
gas. 

If  however  the  Joule  cycle  (Fig.   22)  is   traced    in    the 


Fig.  22. 


counter-clockwise  direction  the  temperatures  cannot  remain 
as  they  were  during  the  motor  cycle.  During  the  motor 
cycle  To  <  Th  and  Ts  =  Th,  heat  passing  from  the  hot  body 
(temperature  Tu)  to  the  gas  (whose  temperature  is  less 
than  Th)  during  the  process  2  3.  But  during  the  purwp 
cycle  heat  must  pass  from  the  gas  to  the  hot  body  during 
the  process  3  2;  this  would  be  impossible  if  Ts  remained 
equal  to  Th  and  Tz<Th-  Thus  during  the  pump  cycle 
T2  must  equal  Th  and  T^  must  be  less  than  Th-  During 
the  motor  cycle  the  gas  reaches  the  temperature  of  the 


I04  THERMODYNAMICS 

hot  body  at  3  while  during  the  pump  cycle  the  gas  reaches 
the  temperature  of  the  hot  body  at  2. 

Exercise  120.  At  what  point  of  the  Joule  cycle,  Fig.  22, 
does  the  gas  attain  the  temperature  of  the  cold  body  (a)  during 
the  motor  cycle,  (b)  during  the  pump  cycle?  Explain  the  passage 
of  the  air  thru  the  engine.  Fig.  23,  page  96,  in  connection 
with  Fig.  22,  when  the  engine  operates  as  a  heat  pump. 

Heat  pumps  are  used  as  refrigerating  machines.  The 
cold  body  from  which  heat  is  withdrawn  is  the  chamber 
to  be  maintained  at  low  temperature  and  the  hot  body  is 
the  cooling  water  which  carries  away  the  heat  rejected 
by  the  gas  at  the  relatively  higher  temperature. 

The  lowest  pressure  in  the  cycle  may  be  atmospheric. 
In  this  case  the  air  of  the  refrigerated  chamber  may  be 
taken  directly  into  the  compression  cylinder  and  the  working 
cylinder  would  discharge  into  the  refrigerated  chamber. 
This  method  is  impractical  on  account  of  the  large  volume 
which  the  required  mass  of  air  occupies  and  also  on  account 
of  the  condensation  and  freezing  of  the  moisture  (carried 
by  the  air  from  the  refrigerated  chamber)  in  the  ports  and 
passages  leading  from  the  working  cylinder.  These  objections 
may  be  overcome  by  working  the  air  in  a  closed  system 
of  pipes  at  a  pressure  above  the  atmospheric  pressure 
thus  decreasing  the  volume  of  the  required  mass  of  air 
and  preventing  the  absorption  of  moisture  by  the  air. 

Refrigerating  machines  working  on  the  principle  just 
described  have  been  largely  superseded  by  machines  using 
liquids  and  their  vapors  instead  of  air.  The  air  machines 
now  are  used  only  on  shipboard  or  other  confined  places 
where  accidents  to  machines  using  ammonia  would  be 
dangerous. 


GAS   CYCLES  I05 

The  action  of  refrigerating  machines  is  judged  by  their 
coefficient  of  performance.  The  thermodynamic  coeffi- 
cient of  performance  of  a  refrigerating  machine  is  defined 
as  the  heat  extracted  from  the  cold  body  divided  by  the 
heat  equivalent  of  the  work  expended  in  driving  the 
machine. 

Referring  to  Fig.  22  the  heat  extracted  from  the  cold 
body  during  the  pump  cycle  is 

mcpiTA—Ti). 

The  heat  equivalent  of  the  work  done  is  the  excess  of  the 
heat  rejected  to  the  hot  body,  mCp{Tz—T2),  over  the  heat 
extracted  from  the  cold  body  mCp{T^—T\) 

thus  mcp\{T^-T-2)-{T^-Tx)\. 

The  coefficient  of  performance  is  thus  under  ideal  con- 
ditions 

iT3-T2)-{n-Ti)     Ts-n  T2_       T2-T1' 

T4-T1     ^     Tx     ^ 

by  reason  of  the  relation  TiT3=Ti7^2- 

From  this  result  it  should  be  noted  that  the  smaller 
the  range  of  temperature  thru  which  the  air  is  cooled  by 
adiabatic  expansion  the  larger  will  be  the  coefficient  of 
performance  and  thus  the  more  economical  the  action  of 
the  machine.  Theoretically  it  is  thus  more  advantageous 
to  cool  the  whole  mass  of  air  to  be  refrigerated  a  few  degrees 
than  to  cool  a  small  portion  thru  many  degrees  and 
then  mix  this  very  cold  air  with  uncooled  air  to  obtain  the 
proper  temperature.      Practically  this  would  require  ve^y 


Io6  THERMODYNAMICS 

large  cylinders  with  large  friction  losses  and  thus  a  reduc- 
tion of  the  actual  coefficient  of  performance.  This  points 
to  one  of  the  advantages  of  the  liquid  and  vapor  machines 
over  the  air  machines  for  refrigeration. 

Exercise  121.  A  refrigerating  machine  is  to  remove  1200 
B.t.u.  per  minute  from  a  cold-storage  room.  The  temperature 
in  this  room  is  to  be  maintained  at  34°  F  under  atmospheric 
pressure.  The  machine  is  to  draw  air  from  the  cold-storage 
room  and  compress  it  to  60  pounds  gage  while  making  100 
working  strokes  per  minute.  The  cooling  water  maintains  the 
coils  at  80°  F.  What  horse-power  is  required  to  drive  this 
machine  under  ideal  conditions  and  what  must  be  the  ideal 
displacement  volumes  of  the  compression  and  of  the  expansion 
pistons? 

The  Warming  Engine. — A  heat  pump  used  to  transfer 
heat  from  the  cold  outside  air  to  the  interior  of  a  building 
may  be  called  a  warming  engine.  Lord  Kelvin  pointed 
out  that  the  methods  now  used  for  warming  buildings 
(by  means  of  stoves  or  radiators)  are  very  wasteful,  at 
least  theoretically.  The  heat  supplied  by  a  stove  at  high 
temperature  is  not  used  to  the  best  advantage  when  it 
is  simply  transferred  to  the  air  of  the  room  to  be  heated 
by  convection  and  radiation.  If  this  heat  at  high  tempera- 
ture be  used  in  a  heat  motor  and  the  energy  so  derived  be 
utilized  to  drive  a  heat  pump  which  transfers  heat  from 
the  cold  outer  air  to  the  building  more  heat  could  thus  be 
delivered  at  the  intermediate  temperature  of  the  building. 

As  a  rough  mechanical  analogy  consider  two  supplies 
of  water  one  at  a  high  level  the  other  at  a  lower  level  and 
assume  that  water  is  required  at  an  intermediate  level. 
Three  ways  of  obtaining  this  supply  suggest  themselves. 


GAS   CYCLES  I07 

Firstly,  the  water  from  the  higher  level  may  simply  be 
allowed  to  fall  to  the  intermediate  level.  Secondly,  the 
water  falling  to  the  intermediate  level  may  be  used  in  a 
turbine  and  the  energy  so  recovered  may  be  utilized  to 
run  a  pump  which  raises  water  from  the  low  to  the  inter- 
mediate level.  Thirdly,  the  whole  fall  of  the  water  from  the 
high  to  the  low  level  may  be  utilized  in  the  turbine  and  the 
energy  so  recovered  may  be  used  to  pump  water  from  the 
low  to  the  intermediate  level. 

Similarly  to  warm  a  building  each  B.t.u.  liberated  by 
the  stove  may  be  transferred  directly  to  the  air  of  the  room 
but  in  this  manner  only  one  B.t.u.  of  heat  will  be  available 
per  B.t.u.  supplied.  Instead  the  B.t.u.  supplied  at  high 
temperature  may  be  partly  transformed  into  mechanical 
energy  by  means  of  a  heat  motor  before  the  remainder  is 
rejected  to  the  air  of  the  room  at  room  temperature.  The 
mechanical  energy  obtained  may  be  used  to  run  a  heat 
pump.  Under  ideal  conditions  this  pump  rejects  as  heat 
to  the  air  at  room  temperature  not  only  the  mechanical 
energy  supplied  to  run  the  pump  but  also  such  heat  as 
it  pumps  from  the  outer  air. 

To  take  a  concrete  example  assume  the  stove  temper- 
ature as  500°  F,  the  temperature  of  the  room  60°  F,  and 
the  outer  air  at  20°  F.  An  ideal  heat  engine  operating 
between    500    and    60°    F   would    have    an    efficiency    of 

^^^  =  o.4=;q.      Thus   for   each   B.t.u.   available   at    ^00°   F, 
960        ^^  ^  ' 

0.459  B.t.u.  would  be  transformed  into  mechanical  energy 

and  0.541  B.t.u.  would  be  rejected  to  the  air  of  the  room 

at  60°  F.     An  ideal  heat  motor  operating  on  a  reversible 

cycle  between   60   and    20°  F  would  have  an  efficiency  of 


Io8  THERMODYNAMICS 

=  0.0769.      It  would  be  very  inefficient.      If  however 

520 

this  motor  be  reversed  and  used  as  a  heat  pump  it  would 
deliver  one  B.t.u.  at  60°  F  for  every  0.0769  B.t.u.  supplied 
to  it  in  the  form  of  mechanical  energy  or  it  would  transfer 
0.9231  B.t.u.  from  the  outer  air  to  the  room  with  an  absorp- 
tion of  mechanical  energy  equivalent  to  0.0769  B.t.u.;   its 

o  92 "? I 
ideal  coefficient  of  performance  would  thus  be         ;;    =12.0. 

0.0769 

As  0.459   B.t.u.  are  available  in  the  form  of  mechanical 

energy  from  each  B.t.u.  furnished  by  the  stove  — — ~—  5.97 
^■^  0.0769 

B.t.u.  will  be  delivered  by  the  heat  pump  to  the  room. 

Therefore  each  B.t.u.  at  500°  F  may  be   made  to  furnish 

5.97+0.541  =  6.51  B.t.u.  at  60°  F,  of  course   under   ideal 

conditions. 

Exercise  122.  In  the  above  example  assume  the  ideal  heat 

motor  to  operate  between  the  stove  and  the  outer  air  as  hot 

and  cold  bodies  respectively  and  the  ideal  heat  pump  to  operate 

between  the  outside  air  and  the  building  as  cold  and  hot  bodies 

respectively.     Compute    the   B.t.u. 's   supplied    to    the    building 

under  these  conditions'  for  each  B.t.u.  furnished  by  the  stove. 

Section  XVII 

CYCLES  OF  INTERNAL-COMBUSTION   ENGINES 

One  of  the  difficulties  met  with  in  the  operation  of  hot-air 
engines  is  the  transfer  of  heat  from  the  furnace  to  the  air 
in  the  cylinder  which  passes  thru  the  processes  required 
to  complete  the  cycle.  To  avoid  this  transfer  of  heat  the 
engine  may  be  designed  so  as  to  allow  combustion  to  occur 
within  the  cylinder  of  the  engine. 


GAS   CYCLES 


109 


The  Brajrton  Cycle. — Brayton  designed  an  engine  in 
which  a  mixture  of  gas  and  air  is  first  compressed  into  a 
receiver,  from  this  receiver  it  flows  (thru  wire  gauze 
to  prevent  backfire)  into  the  working  cyHnder  where  the 
mixture  is  ignited.  During  this  process  the  pressure  in 
the  cyHnder  must  be  the  same  as  the  pressure  in  the  re- 
ceiver. After  the  flow  of  the  mixture  is  cut  off  the  heated 
products  of  combustion  expand  adiabatically,  or  nearly 
so,  to  atmospheric  pressure  provided  the  cyHnder  volume 
is  sufficiently  large.  Finally  the  expanded  products  of 
combustion  are  expelled  to  the  atmosphere  at  constant 
back  pressure. 

Exercise  123.  Sketch  the  />D-diagram  of  the  Brayton  cycle 
and  note  that  it  is  thermodynamically  equivalent  to  the  Joule 
cycle,  see  page  94. 

The  Otto  Cycle. — An  engine  operating  on  the  Otto 
cycle  requires  only  one  cylinder.  The  cycle  is  completed 
in  four  strokes  of  the  piston.  In  Fig.  24  the  processes  are 
indicated  as  follows: 

(i)  The  explosive  mixture  is  drawn  into  the  cylinder, 
process  i  2. 

(2)  This  mixture  is  compressed,  process  2  3. 

(3)  The  compressed  charge  is  ignited,  the  combustion 
assumed  to  be  instantaneous  causes  an  increase  of  pressure 
and  temperature  at  constant  volume,  process  3  4. 

(4)  The  products  of  combustion  expand,  process  4  5. 

(5)  The  exhaust  valve  opens  permitting  the  gases  to 
escape  to  the  atmosphere,  process  5  2. 

(6)  The  remaining  products  of  combustion  are  now 
expelled  into  the  atmosphere,  process  2  i. 


no  THERMODYNAtilJo 

Thermodynamically  the  Otto  cycle  is  represented  by 
23452.  The  process  5  2  is  regarded  as  a  cooHng  at  con- 
stant volume  of  the  whole  mass  of  gas  taken  in  as  the 
charge.  This  mass  of  gas  is  then  conceived  to  be  recom- 
pressed,  2  3,  and  reheated,  3  4,  etc.,  altho  actually  a  new 
charge  must  be  taken  in.  Moreover  in  gas-engine  design 
no  account  is  usuallv  taken  of  the  fact  that  the  contents 


Fig.  24. 


of  the  cylinder  is  not  pure  air.     Calculations  made  in  this 
way  give  results  which  are  known  as  the  air  standard. 

Analysis  of  the  Otto  Cycle. — Let  T^3=T^4=Fh  and 
V2=V5=Vc  and  assume  the  processes  2  3  and  4  5  (Fig. 
24)  to  be  polytropic. 

p2        \Vh/  pb 


Then 


As 


we  have 


pv 

T  '' 


■a  constant, 


Tz  T4.         n  T2 


GAS   CYCLES  III 

Thus  ^  =  ^=1?  =  ^ 

p4.      po       To       14: 

see  also  page  loi. 

The  amounts  of  heat  supplied  during  the  various  processes 
are 

2Q3  =  mCn(T3—T2) 
3QA=mCv{T4—T3) 

aQo  =  mCniTo—  T4)  =  —  mCu(T4—  T5) 
5Q2  =  )nCv(T2—To)  =  —mCviT-o—  T2), 

H  —  k 

where  Cn= Cv. 

n—i 

The  work  done  by  the  gas  during  the  various  processes  is 

jjr                P2VC—P3VH 
214/3  =  

n—i 

n—i 

5W2  =  o 

and  the  net  work  of  the  cycle  is 

j^,  _(P^-p3)VH-(p5-p2)Vc 


As 


n—i 


,PK3=^(^.-A3)ji'.-({;^)"iv| 

n-i\P3     Vr      V'c 


112  THERMODYNAMICS 

Exercise  124.  Show  that  the  above  result  is  equivalent  to 


'-'--mr)m 


n-l 


Exercise  125.  Show  that  the  mean  effective  pressure  of  the 
Otto  cycle  is 

n-l  1 

2W2  ^    /pi  \  \p2/  ^  \p2 


(^^) 


Vc—Vh      '\p3      I       n—i 


i^y- 


In  gas-engine  design  the  calculations  are  usually  made 
under  the  assumption  that  the  compression  and  the  expan- 
sion are  adiabatic.  Under  these  conditions  the  net  work 
of  the  cycle  may  be  expressed  as  follows: 

2W2  =  J(3Qi-\-5Q2)=JmC„(Ti-T3+T2-T5). 

Exercise  126.  Show  that  the  efficiency  of  the  Otto  cycle  is 

t-i 
T2  T,  /V„Y-'  /pA    * 


when  compression  and  expansion  are  adiabatic. 

Exercise  127.  (a)  Show  that  the  efficiency  of  the  Otto  cycle 
is  less  than  the  efficiency  of  the  Carnot  cycle  both  cycles  operating 
between  the  same  hot  and  the  same  cold  bodies. 

(b)  Show  that  the  efficiency  of  the  Otto  cycle  increases  with 
increasing  compression. 

The  Diesel  Cycle. — The  Diesel  cycle  is  particularly 
suited  to  the  internal  combustion  of  oil.     In  this  cycle, 


GAS   CYCLES 


"3 


Fig.  25,  the  cylinder  takes  in  a  charge  of  air  which  is  then 
compressed  adiabatically,  process  i  2,  to  about  500  pounds 
per  square  inch  into  the  clearance  space.  The  temperature 
of  the  compressed  air  is  so  high  that  the  oil  injected  into 
the  cylinder  ignites  spontaneously.  The  injection  of  oil 
and  the  heating  of  the  contents  of  the  cylinder  continues 
during  a  portion  of  the  working  stroke  of  the  piston.  After 
the  injection  of  oil  ceases  the  contents  of  the  cylinder 
expand  adiabatically  to  the  end  of  the  stroke. 


Fig.  25. 

Theoretically  at  least  the  flow  of  oil  may  be  regulated  so 
as  to  maintain  either 

(a)  constant  pressure  until  the  oil  is  cut  ofT,  process  2  3, 

(b)  constant  pressure  to  A  and  then  constant  temperature 

to  B,  process  2  A  B, 

(c)  constant   temperature   thruout   the  period  of  injection 

of  oil,  process  2  C. 
Thermodynamically    the    cycle    is   completed    after    the 
adiabatic  expansion  34,  54,  or  C  4  by  cooling  at  constant 
volume  from  4  to  i,  practically  the  products  of  combustion 
are  expelled  to  the  atmosphere. 


114  THERMODYNAMICS 

Exercise  128.  Show  that  the  efficiency  of  the  Diesel  cycle 
under  the  conditions  described  above  is 


c.(r.-r,)-..(r.-r,)^__,/r^\  ,^^^„_,,,,,_,_^,„,_ 


Ci>{Ts-T2)  k\T',-T, 


-— ,  for  condition  (b), 
kiT^-T,)+{k-i)TAlog-^ 

V  A 

I  Vc  \  (VcY-'^      1 

jpiVilog- — mcv{Ti-Ti)  CvTiU~]       -i 

I  — 


I  Vc  R  Vc         ' 

for  condition  (c). 

Exercise  129.  How  many  B.t.u.  per  i.h.p.  per  hour  must 
be  supplied  to  an  ideal  engine  operating  on  an  Otto  cycle? 

Exercise  130.  Compute  the  ideal  efficiency  of  the  Otto  cycle 
when  compression  is  carried  to  1 5  atmospheres. 

Exercise  131.  It  is  claimed  that  an  oil  engine  uses  0.42 
pound  of  fuel  (18,000  B.t.u.  per  pound)  per  kilowatt-hour 
delivered  by  the  generator  to  which  it  is  directly  connected, 
that  20  per  cent  of  the  indicated  power  is  lost  between  the 
cylinder  and  the  bus  bars,  that  the  highest  temperature  in  the 
cylinder  does  not  exceed  2500°  F,  and  that  the  exhaust  temper- 
ature is  800°  F.     Are  the  above  claims  probable? 


CHAPTER  VI 
THE   SECOND   LAW  OF  THERMODYNAMICS 

Section  XVIII 
THE   FIRST   LAW 

Conservation  of  Energy. — Experience  and  experimen- 
tation have  always  led  to  the  conclusion  that  the  total 
energy  in  a  given  space  cannot  be  increased  or  diminished 
unless  energy  passes  thru  the  boundary  of  the  space 
considered.  This  principle  is  known  as  the  law  of  conser- 
vation of  energy.  In  accordance  with  this  law  the  total 
energy  of  the  universe  is  a  fixed  unalterable  quantity. 

Transformation  of  Energy. — Altho  the  total  energy 
within  a  given  space  cannot  be  increased  or  diminished 
without  introducing  energy  into  or  withdrawing  energy 
from  the  given  space,  energy  of  one  form  may  be  transformed 
into  energy  of  another  form  within  this  space  without  pro- 
ducing changes  in  the  amount  of  energy  residing  outside 
of  the  space  considered.  Thus  mechanical  energy  may  be 
converted  into  electrical  energy  and  under  ideal  conditions 
it  is  possible  to  conceive  that  all  of  the  mechanical  energy 
which  disappears  would  reappear  as  electrical  energy. 
Practically  this  complete  conversion  is  impossible.  Unavoid- 
able losses  always  result  in  the  transformation  of  some  of 
the  mechanical  energy  into  other  forms  of  energy,  principally 

"5 


Il6  THERMODYNAMICS 

heat,  with  a  corresponding  decrease  in  the  amount  of  electrical 
energy  produced. 

The  First  Law  of  Thermodynamics  states  that  whenever 
heat  is  transformed  into  mechanical  energy  or  mechanical 
energy  is  transformed  into  heat  a  definite  quantity  of 
mechanical  energy  reappears  for  every  definite  quantity 
of  heat  which  disappears  and  vice  versa.  It  has  been  ex- 
perimentally determined  that  the  complete  transformation 
of  777.64  foot-pounds  of  mechanical  energy  always  results 
in  the  appearance  of  i  B.t.u.  of  heat. 

It  should  be  noted  that  the  first  law  of  thermodynamics 
does  not  specify  how  much  of  a  given  quantity  of  heat 
energy  can  be  transformed  into  mechanical  energy.  It 
simply  states  that  if  any  portion  of,  or  if  the  whole  of  any 
given  quantity  of  heat  is  transformed  each  heat  unit  that 
disappears  will  reappear  as  a  definite  number  of  units  of 
mechanical  energy. 

Section  XIX 
THE   SECOND   LAW   OF   THERMODYNAMICS 

This  law  has  been  stated  in  many  ways.  As  the  proof 
of  this  second  law  must  be  based  upon  experience  and  ex- 
perimentation the  simplest  form  of  the  law  is  the  following : 
Heat  has  never  been  known  to  flow  of  its  own  accord 
from  a  cold  to  a  relatively  hotter  body;  that  is,  an  ex- 
penditure of  energy  is  always  necessary  to  cause  heat  to 
flow  from  a  body  to  another  body  of  higher  temperature. 

The  consequences  of  this  law  are  of  the  greatest  importance. 
It  will  be  shown  by  means  of  this  second  law  that  heat 
can  never  be  completely  converted  into  mechanical  energy, 
not  even  under  ideal  conditions;  that  the  fraction  which 


THE    SECOND   LAW   OF   THERMODYNAMICS  II7 

is  convertible  depends  upon  the  available  extreme  tem- 
peratures; that  this  fraction  depends  only  upon  these 
temperatures  and  not  upon  the  working  substance  with 
which  the  heat  is  associated ;  and  thus  that  every  transforma- 
tion (even  under  ideal  conditions)  of  heat  into  mechanical 
energy  is  accompanied  by  an  unavoidable  waste  of  some 
of  the  heat  energy  involved. 

In  order  to  demonstrate  these  consequences  of  the  second 
law  we  must  first  carefully  consider  the  nature  of  reversible 
and  irreversible  processes  and  cycles. 

Reversible  Processes  and  Cycles. — A  process  is  said  to 
be  reversible  when  the  changes  involved  may  be  retraced 
in  reverse  order,  thus  returning  the  substance  undergoing 
change  to  its  initial  state,  and  at  the  same  time  leave  no 
change  of  any  kind  in  any  associated  bodies. 

As  an  example,  during  an  isothermal  expansion  of  ideal 
gases  not  only  does  the  state  of  the  gas  change  from  pi, 
Vi,  Ti  to  p2,  Vo,  To  but  heat  is  supplied  to  the  gas  during 
the  change  and  the  work  performed  by  the  gas  must  be 
stored  in  some  way.  Thus  two  bodies  associated  with 
the  gas  during  the  change  of  state  were  altered  one  by 
being  deprived  of  some  of  its  heat  and  the  other  by  having 
its  store  of  mechanical  energy  increased. 

If  this  isothermal  process  is  reversed  and  the  gas  is 
compressed  to  its  initial  state,  work  must  be  performed 
on  the  gas  and  heat  must  be  withdrawn  from  the  gas. 
The  necessary  work  may  be  obtained  from  the  body  in 
which  the  work  performed  during  expansion  was  stored. 
The  heat  withdrawn  from  the  gas  may  be  returned  to  the 
body  which  acted  as  the  original  source  of  heat.  Then 
after  the  gas  has  been  returned  to  its  initial  state  no  changes 


Il8  THERMODYNAMICS 

will  remain  in  any  bodies  associated  with  the  gas  during 
its  changes  of  state,  because  all  work  performed  during 
expansion  must  be  used  for  compression  and  all  heat  bor- 
rowed during  expansion  will  be  returned  during  compression. 

It  is  at  once  evident  that  the  processes  just  described 
are  ideal.  Practically  it  is  impossible  to  realize  storage 
and  redelivery  of  the  work  involved  without  some  loss  due 
to  friction.  Nor  can  heat  be  transferred  from  the  source 
to  the  gas  while  both  are  at  the  same  temperature  or  while 
they  differ  in  temperature  by  an  infinitesimal  amount  as 
they  must  during  a  reversible  change  of  state.  Practical, 
actually  realizable  processes  are  never  reversible. 

Under  ideal  conditions  processes  may  be  conceived  to 
be  reversible  and  any  cycle  composed  wholly  of  reversible 
processes  must  also  be  reversible  in  the  thermodynamic 
sense.  Thus  the  Carnot  cycle  is  reversible.  The  Stirling 
and  the  Ericsson  cycles  with  perfect  regeneration  are  also 
reversible.  Without  regeneration  the  isopiestic  processes 
of  the  Stirling  cycle  and  the  isometric  processes  of  the 
Ericsson  cycle  are  irreversible.  During  these  processes 
the  gas  is  not  at  the  same  temperature  as  the  body  supply- 
ing or  receiving  heat  and  a  reversal  of  the  process  would 
require  the  flow  of  heat  from  a  cold  gas  to  a  relatively  hotter 
body. 

It  must  not  be  supposed  that  the  Stirling  cycle  without 
regeneration  cannot  be  traced  in  a  counter-clockwise  direc- 
tion and  the  cycle  used  in  a  heat  pump.  This  can  be  done 
but  only  after  the  conditions  of  operation  have  been  changed 
so  that  during  the  isometric  processes  the  thermal  con- 
tact between  the  gas  and  the  hot  and  the  cold  bodies 
are  interchanged. 


THE    SECOND   LAW   OF   THERMODYNAMICS  II9 

Exercise  132.  Sketch  a  Stirling  cycle  without  regeneration 
and  indicate  during  which  processes  the  gas  is  in  thermal  contact 
with  the  hot  body  and  with  the  cold  body  during  (a)  the  motor 
cycle,  {b)  the  pump  cycle.  How  do  these  conditions  differ 
from  the  conditions  existing  when  a  regenerator  is  used? 

The  Joule  cycle  is  also  irreversible,  see  page  103.  Any 
process  involving  the  transfer  of  heat  from  one  body  to 
another  at  an  appreciably  lower  temperature  is  irreversible. 

As  conduction  of  heat  is  always  present  in  all  actual 
machines  reversibility  can  never  be  attained  in  actual 
practice.  It  is  an  ideal  condition  which  may  be  imagined 
but  not  realized.  The  assumption  of  ideal  reversible 
processes  and  cycles  is  very  useful  in  arriving  at  important 
deductions  of  the  second  law  of  thermodynamics. 

Irreversible  Processes  and  Cycles. — xA.s  another  example 
of  an  irreversible  process  consider  the  flow  of  air  in  Joule's 
experiment  devised  to  show  that  the  change  in  the  internal 
potential  energy  of  a  gas  is  zero,  see  page  10.  This 
change  of  state  involves  conditions  which  have  not  yet 
been  studied.  The  internal  energy  of  the  compressed  gas 
is  partly  changed  to  kinetic  energy  of  the  moving  mass  of 
gas  during  its  passage  from  the  high-pressure  to  the  low- 
pressure  receiver.  This  kinetic  energy  is  then  reconverted 
into  heat  energy  in  the  low'-pressure  receiver  thru 
impact  and  internal  friction.  The  initial  and  the  final 
states  of  the  gas  are  readily  shown  on  the  pV -plane  but 
the  intermediate  states  cannot  be  represented  by  an  adia- 
batic  pV^  =  a.  constant  even  tho  no  heat  is  supplied  to 
or  withdrawn  from  the  gas  during  the  process  because  the 
whole  mass  of  gas  is  not  always  under  the  same  pressure 
and  temperature. 


I20  TIIEJIMODYNAMICS 

That  this  process  is  not  a  reversible  adiabatic  process 
such  as  has  been  studied  becomes  evident  when  we  attempt 
to  restore  the  gas  to  its  initial  condition  without  leaving 
changes  of  any  kind  in  any  associated  bodies.  The  gas 
will  of  course  not  flow  back  into  the  high-pressure  receiver 
of  its  own  accord.  It  must  be  pumped  back.  Assume 
the  compression  to  be  isothermal  then  work  must  be  sup- 
plied and  the  body  supplying  this  work  will  lose  this  energy. 
Assume  the  compression  to  be  reversibly  adiabatic  then 
again  must  work  be  supplied  and  more  work  than  during 
isothermal  compression.  Under  the  last  assumption  the 
compressed  gas  will  be  hotter  than  in  its  initial  state.  It 
may  be  argued  that  this  extra  heat  may  be  withdrawn 
from  the  compressed  gas,  transformed  into  work,  and  this 
work  used  to  repay  the  loan  of  work  made  for  compress- 
ing the  gas.  It  will  be  shown  that  the  work  so  obtained 
will  be  insufficient  to  fully  repay  the  loan  of  energy. 
Under  any  conceivable  conditions  return  to  the  initial  state 
is  only  possible  when  changes  of  some  kind  remain  in 
associated  bodies.  If  this  is  so  then  the  process  is  irre- 
versible. 

Carnot's  Principle. — This  principle  is  a  consequence  of 
the  second  law  of  thermodynamics.  It  may  be  stated  as 
follows : 

No  heat  motor  can  have  a  greater  efficiency  than  the 
efficiency  of  a  heat  motor  operating  on  a  reversible  cycle 
provided  all  motors  operate  between  the  same  extreme 
temperatures. 

To  prove  this  principle  conceive  two  heat  motors  (marked 
I  and  2,  Fig.  26)  operating  between  the  same  hot  and 
the  same  cold  body.     Let  motor  i  operate  on  a  reversible 


THE    SECOND   LAW    OF   THERMODYNAMICS 


121 


cycle  and  assume  that  the  efficiency  of  motor  2  is  greater 
than  the  efficiency  of  motor  i,  so  that, 

'72>r;i. 

If  the  principle  we   are   to  prove  is  true  this  assumption 
should  lead  to  an  impossible  condition. 

As  shown  in  Fig.  26  when  both  motors   are  running  as 
motors  we  may  assume  conditions  such   that  each  motor 


Hot  Body 


/Q 


Q^ 


I — - — [  ^iQ<'^2Q 


Cold  Body 

Fig.  26. 


not  Body 


1 

« —     ■* — 

2 

Cold  Body 
Fig.  27. 


draws  the  same  quantity  of  heat  from  the  source  of  heat 
in  the  same  time.  Under  the  assumption  r72>r/i  the  heat 
transformed  into  mechanical  energy  by  motor  2  must  be 
greater  than  the  mechanical  energy  delivered  by  motor 
I  or 

ViQ<V2Q. 

Assume  now  that  motor  i  is  operated  as  a  heat  pump, 
Fig.  27,  and  that  the  whole  mechanical  energy  obtained 
from  motor  2  is  used  to  drive  the  reversed  motor  i.  As 
this  motor  is  able  to  transform  into  work  771  B.t.u.  per  B.t.u. 
supplied   when  running   as  a  motor  it  will  deliver  to  the 


12  2  THERMODYNAMICS 

source  of  heat  one  B.t.u.  for  every  771  B.t.u.  supplied  to  it 
as  work  when  it  runs  as  a  heat  pump.  Under  our  assump- 
tion motor  I  will  receive  772^  B.t.u.  of  mechanical  energy 

and  it  must  therefore  reject  to  the   hot  body  -^  B.t.u. 

m 

But  772  >  7?!  by  hypothesis.  Therefore  during  the  operation 
of  the  motors  as  shown  in  Fig.  27  more  heat  will  be  re- 
turned in  a  given  time  to  the  hot  body  by  reversed  motor 
I  than  motor  2  extracts  from  the  hot  body.  Thus  we  have 
a  self-contained  system  in  which  heat  continuously  flows 
from  a  cold  to  a  relatively  hotter  body,  an  impossibility 
according  to  the  second  law  of  thermodynamics.  We  thus 
conclude  that  our  hypothesis  is  impossible  and  that  772 
cannot  be  greater  than  771. 

Exercise  133.  Establish  the  inequalities  indicated  in  Fig.  27. 

The  Consequences  of  the  Second  Law  of  Thermo- 
dynamics.— As  has  just  been  shown  any  heat  motor  opera- 
ting on  a  reversible  cycle  has  an  efficiency  at  least  as  high 
as  any  other  heat  motor  (reversible  or  not)  utilizing  the 
same  hot  and  cold  bodies  as  source  and  as  receiver  of  heat. 

Therefore  all  heat  motors  operating  on  reversible  cycles 
must  have  the  same  efficiency. 

Also,  with  a  given  hot  body  and  a  given  cold  body  any 
heat  motor  operating  on  a  reversible  cycle  is  as  efficient 
as  any  heat  motor  can  possibly  be,  or  no  engine  can  convert 
a  greater  fraction  of  the  heat  energy  supplied  to  it  into 
mechanical  energy  than  an  engine  operating  on  an  ideal 
reversible  cycle. 

It  should  be  noted  that  in  the  preceding  discussion  no 
mention  is  made  of  the  working  substance  nor  of  the  mode 


THE   SECOND   LAW   OF   THERMODYNAMICS  1 23 

of  action  of  the  motor.  Reversibility  in  the  thermody- 
namic sense  is  the  only  stipulation.  Thus  any  working  sub- 
stance passing  thru  any  reversible  cycle  will  transform 
into  mechanical  energy  the  greatest  possible  fraction  of 
the  heat  supplied  for  any  given  conditions  as  regards 
the  source  and  the  receiver  of  heat. 

We  must  now  establish  the  efficiency  of  a  reversible 
cycle  in  terms  of  the  temperatures  of  the  source  and  of  the 
receiver  of  heat  upon  which  alone  this  efl&ciency  depends. 

Section  XX 
KELVIN'S   ABSOLUTE    SCALE    OF   TEMPERATURE 

Lord  Kelvin  established  a  scale  of  temperature  which 
is  independent  of  any  particular  substance  and  is  in  this 
sense  absolute.  That  such  a  scale  of  temperature  is  nec- 
essary at  least  theoretically  becomes  evident  when  we 
remember  that  the  readings  of  mercury  thermometers 
differ  with  each  other  and  with  the  readings  of  an  air  ther- 
mometer and  that  none  of  these  agree  with  the  reading  of 
the  standard  hydrogen  thermometer. 

Kelvin  arbitrarily  defmed  temperature  by  means  of  the 
relation 

6     Q' 

where  6'  and  9  are  the  temperatures  of  a  source  and  of 
a  receiver  of  heat  respectively  and  Q'  and  Q  are  the  quan- 
tities of  heat  supplied  by  this  source  and  rejected  to  this 
receiver  by  any  heat  motor  operating  on  a  reversible  cycle 
between  them. 


124 


THERMODYNAMICS 


As  the  efficiency  of  any  heat  motor  is  by  definition 
V         ^     , 

Q'    Q'         Q    Q' 

and  as         -:r=^    or     ^  =  ^=r,  some  constant, 


we  have 
so  that 


Q  G 

Q'^rd'     and     Q  =  rd, 


!  =  ■ 


Fig.  28. 

This  is  the  efficiency  of  any  and  every  ideal  reversible 
cycle  and  it  cannot  be  exceeded  by  any  other  cycle  oper- 
ating between  the  same  source  and  the  same  receiver  of 
heat.  This  expression  is  as  yet  meaningless.  The  sig- 
nificance of  d  must  be  established.  The  scale  of  tempera- 
ture must  be  determined. 

The  Carnot  cycle  is  a  typical  reversible  cycle.  As  shown 
in   Fig.    28   this   cycle   may  be    represented    by   any   two 


THE   SECOND   LAW   OF   THERMODYNAMICS  125 

adiabatics  and  two  isothermals.  Let  the  cycle  marked 
/  be  the  cycle  of  a  heat  motor  operating  between  the  tem- 
peratures di  and  6-2  and  let  Qi  be  the  heat  supplied  at  a 
temperature  ^i  and  Q2  be  the  heat  rejected  by  the  motor 
at  a  temperature  do.  Further  assume  that  the  heat  so 
rejected  by  motor  /  is  at  once  supplied  to  another  motor 
(whose  cycle  is  marked  II)  at  the  temperature  62,  and 
that  this  motor  rejects  Qs  heat  units  at  a  temperature 
ds,  and  so  on. 

Under  these  conditions  the  efficiency  of  motor  /  is 

Q1-Q2     9,-02 

and  for  motor  // 

Q2  —  Qs     02  —  03 

..  ''=~Qr=~0r' 

But,  by  definition, 

h  =  Ql     or     2i  =  22 

02      Q2  01        02 

and  as  0i,  02,  03  are  to  be  consecutive  points  on  a  tempera- 
ture scale  the  interval  between  0i  and  02  must  equal  the 
interval  between  62  and  03  or 


Now  from  the  expression  for  rji  and  r/2  we  find 

Ql-Q2  =  ^{01-02) 

and  Q2-Q3  =  ^{02-03) 

whence  by  reason  of  the  above    relations   between  the  ^'s 
it  follows  that 

Qi-Q2  =  Q2-Q3. 


126  THERMODYNAMICS 

Therefore  the  heat  which  disappears  and  which  is  con- 
verted into  mechanical  energy  is  the  same  in  each  of  the 
ideal  reversible  motors  when  the  temperatures  of  their 
sources  and  receivers  of  heat  differ  by  the  same  number 
of  degrees  on  Kelvin's  absolute  scale  of  temperature.  Thus 
having  assumed  any  two  isothermals,  Fig.  28,  at  say  one 
degree  apart  the  next  isothermal  another  degree  lower 
in  temperature  must  be  so  located  that  the  area  of  its 
cycle  (marked  //)  must  equal  the  area  of  the  first  cycle 
(marked  /). 

As  each  succeeding  motor  transforms  a  portion  of  the 
heat  originally  supplied  to  the  first  motor  into  mechanical 
energy  and  as  the  original  supply  of  energy  need  not  be 
infinite  and  as  each  motor  transforms  the  same  amount  of 
heat  as  the  preceding  one,  the  heat  will  finally  be  wholly 
transformed.  When  this  occurs  the  last  motor  must 
operate  with  a  receiver  at  zero  temperature  and  this  will 
be  an  absolute  zero.  There  is  nothing  beyond.  The  zero 
of  Kelvin's  scale  is  an  absolute  zero. 

To  put  it  in  another  way  assume  that  in 


77  = 


d  becomes  negative,  a  negative  temperature  on  Kelvin's 
scale,  then  7;  becomes  greater  than  one  or  the  motor  would 
deliver  more  mechanical  energy  than  would  be  equivalent 
to  the  heat  supplied  to  it.  This  would  violate  the 
law  of  conservation  of  energy  and  our  assumption  is 
impossible. 

To   establish    a    definite    numerical   value    for   Kelvin's 
scale  of  temperature  assume    Oi    to    correspond  to  212°  F 


THE   SECOND   LAW   OF   THERMODYNAMICS  1 27 

and  02  to  32°  F.  Between  di  and  62  we  have  the  follow- 
ing relations 

6-2      Q2 
and  01 -02  =180. 

Next  assume  a  working  substance  which  passes  thru 
the  ideal  Carnot  cycle.  Any  substance  will  do  for  all 
substances  have  the  same  efficiency.  Let  the  substance 
be  air.     Now  find  by  experimentation  and  calculation  the 

value  of  ^.*     Its  value  is 
Q2 


Q2 

Then  as 

^1        ^1                  AA 
-  =  ^-=1.3663 

and 

01 -02  =180, 

we  find  that 

02  =  492°  F. 

Kelvin's  scale  for  practical  purposes  is  the  same  as  the 
absolute  scale  already  established  on  page  12. 

The    efficiency   of   all   reversible    cycles    is    therefore 


V  =  ^ — 


where  Th  is  the  absolute  temperature  of  the  source  of 
heat  and  Tc  is  the  absolute  temperature  of  the  receiver 
of  heat. 

*  Callendar:  Phil.  Mag.  (b)  5,  48.     1903. 


128  THERMODYNAMICS 

Section  XXI 
THE   AVAILABILITY   OF   HEAT   ENERGY 

The  availability  of  heat  energy  is  measured  by  the  frac- 
tion of  the  heat  energy  which  can  be  transformed  into 
mechanical  energy  under  the  best  conceivable  conditions. 
Thus  the  efficiency  of  any  reversible  cycle  operating  be- 
tween the  temperature  of  the  given  heat  energy  and  the 
lowest  obtainable  temperature,  i.e.  the  temperature  of  the 
coldest  body  which  can  be  relied  upon  continuously  to 
receive  heat  without  increasing  its  temperature  appreci- 
ably, is  a  measure  of  the  availability  of  the  heat  supplied. 

Thus  never,  even  under  the  most  ideal  conditions  which 
may  be  conceived  altho  not  realized,  can  any  given 
supply  of  heat  be  completely  transformed  into  mechanical 
energy,  for  this  would  only  become  possible  if  a  receiver 
of  heat  could  be  found  whose  temperature  would  remain 
permanently  at  absolute  zero. 

The  degradation  of  heat  energy  is  the  term  used  to 
denote  the  idea  that  every  transformation  of  heat  energy 
into  mechanical  energy  must  be  accompanied  by  an  irre- 
trievable loss,  not  of  energy,  but  of  transformable  energy. 
All  heat  rejected  to  a  receiver  of  heat  whose  temperature 
is  the  lowest  obtainable  is  degraded;  it  can  never  be  even 
partly  transformed  into  mechanical  energy.  Such  heat 
may  be  used  for  heating  the  receiver  of  heat  but  not  for 
the  production  of  mechanical  energy. 

The  availability  of  heat  energy  for  transformation  into 
mechanical  energy  thus  depends  upon  the  fall  in  tempera- 
ture between  the  source  and  the  receiver  of  lowest  obtainable 


THE    SECOND    LAW   OF    THERMODYNAMICS  1 29 

temperature.  It  is  measured  by  the  efficiency  of  an  ideal 
reversible  cycle  operating  between  the  source  and  this 
receiver  of  heat. 

Altho  we  can  never  attain  this  efficiency, 

_Th-Tc 
V— — ^ — , 

it  is  the  standard  by  which  all  actual  thermal  efficiencies 
should  be  judged. 

Exercise  134.  What  percentage  of  the  heat  suppUed  by  a 
source  at  600°  F  must  necessarily  be  wasted  if  the  lowest  ob- 
tainable receiver  temperature  is  60°  F? 

Exercise  135.  (a)  The  combustion  of  gases  furnishes  a  source 
of  heat  with  a  temperature  of  2500°  F.  Assuming  that  the 
lowest  available  temperature  is  80°  F  and  that  the  heat  is 
supphed  to  a  boiler  and  transmitted  to  steam  at  380°  F  before 
being  suppUed  to  an  engine,  what  is  the  best  possible  thermal 
efl&ciency  of  this  engine? 

(b)  What  would  be  the  efficiency  of  an  ideal  engine  capable 
of  transforming  the  heat  directly  from  the  source? 

Exercise  136.  A  steam  engine  requires  15,000  B.t.u.  per 
i.h.p.  hour.  The  temperature  of  the  steam  supphed  to  this 
engine  is  300°  F,  and  the  temperature  in  the  condenser  is  130° 
F.  What  is  the  ratio  of  the  thermal  efficiency  of  this  engine 
to  the  thermal  efficiency  of  the  best  conceivable  engine  oper- 
ating under  the  same  conditions? 

Three  Types  of  "Perpetual  Motion." — Peqoetual  motion 
of  the  first  type  would  be  realized  by  a  machine  which 
could  create  energy,  a  machine  which  gives  us  something 
for  nothing.  Any  machine  of  this  kind  would  violate  the 
law  of  conservation  of  energy  and  is  therefore  impossible. 


I30 


THERMODYNAMICS 


Perpetual  motion  of  the  second  type  is  sought  by  in- 
ventors who  attempt  to  design  machines  which  are  able 
to  transform  all  of  the  heat  energy  supplied  into  mechan- 
ical energy  without  having  a  receiver  at  a  temperature 
of  absolute  zero.  In  this  case  no  attempt  is  made  to  obtain 
something  for  nothing  and  still  the  result  is  impossible 
of  attainment  for  it  would  mean  a  violation  of  Carnot's 
principle  and  thus  of  the  second  law  of  thermodynamics. 

Perpetual  motion  of  the  third  type  is  embodied  in  an 
ideal  mechanism  which  when  once  set  in  motion  would 
continue  in  motion  forever.  A  mechanism  of  this  kind  is 
not  supposed  to  deliver  energy  and  altho  it  is  not  theo- 
retically impossible  it  could  never  actually  be  constructed. 


CHAPTER  VII 
ENTROPY 

Section  XXII 
INTRODUCTION 

At  the  outset  it  should  be  understood  that  no  attempt 
is  to  be  made  in  the  following  to  explain  what  entropy  is. 
Such  matter  would  be  out  of  place  in  a  first  course  in 
thermodynamics.  Nor  is  it  necessary.  We  do  not  know 
what  electricity  or  even  gravitation  is.  It  will  be  suffi- 
cient if  we  study  the  properties  of  and  become  familiar 
with  entropy  and  above  all  learn  to  recognize  its  usefulness 
in  our  calculations. 

The  availability  of  Q  heat  units  supplied  at  a  tempera- 
ture Th  when  the  lowest  obtainable   receiver  temperature 

is  Tc  is 

^Th-Tc^    _Tc 
"^         Th         '     Th 

During  this  ideal  transformation  of  heat  into  mechanical 
energy  Q{i—ri)  heat  units  must  be  wasted  or  degraded, 

and  Q{^-r^)  =  Q^=I^^^Tc. 

The  quant  it  V  ~  is  called  the  change  in  the  entropy  of 

the  source  due  to  its  loss  of  Q  heat  units  at  a  constant 
temperature  Th- 

131 


132 


THERMODYNAMICS 


The  heat  degraded  during  this  ideal  transformation  of 
heat    into    mechanical   energy   may   therefore  be   said   to 

equal  the  entropy  lost  by  the  source  (^)  multiplied  by 

the  lowest  obtainable  receiver  temperature  Tq. 

Whenever  a  body  rejects  heat  its  entropy  is  said  to 
diminish,  whenever  a  body  receives  heat  its  entropy  is 
said  to  increase.  The  change  in  the  entropy  of  a  body  is 
measured   by   the   heat   received   (or  rejected)   divided  by 


V 


Fig.  29. 

the  absolute  temperature  of  the  body  during  this  transfer 
of  heat. 

Let  us  study  the  changes  of  entropy  which  occur  during 
the  transfers  and  the  transformation  of  heat  during  the 
operation  of  an  ideal  Carnot  cycle,  Fig.  29. 

During  the  process  i  2  assume  that  the  hot  body  rejects 
and  the  working  substance  receives  Qh  heat  units  at  an 
absolute  temperature   Th-     Then  the  entropy  of  the  hot 

body  is  diminished  by  ^  and  the  entropy  of  the  working 


ENTROPY 


133 


substance  is  increased  by  ^.      Similarly  if  the  working 

substance   rejects   Qc  heat   units   to   the   cold   body   at   a 
temperature  Tc  the  entropy  of   the  working    substance  is 

thereby  diminished  by  ^  while  the  entropy   of  the  cold 


Qc  T, 


=  I- 


c 


body  is  increased  by  ^. 
■i  c 

But  77=1  — 

^  Qh  Tu' 

SO  that  ^=% 

see  also  page  123. 

As  no  heat  is  transferred  during  the  processes  2  3  and  4  i 
no  change  in  the  entropy  of  either  the  working  substance 
or  of  any  associated  body  occurs.  Finally  during  the 
operation  of  the  cycle  as  a  whole  the  entropy  of  the  work- 
ing substance  returns  periodically  to  its  initial  value  and 
altho  the  entropy  of  the  hot  body  continually  dimin- 
ishes and  that  of  the  cold  body  continually  increases  the 
entropy  of  the  whole  system  remains  constant. 

Consider  now  a  Carnot  cycle  in  which  owing  to  con- 
duction the  temperature  of  the  working  substance  during 
the  process  i  2  (Fig.  29)  is  Ti,  a  constant,  where  Ti<Ta 
and  the  temperature  along  the  process  3  4  is  To  where 
T2>Tc.  Then  if  Qh  is  the  heat  supplied  by  the  hot 
body  and  Q2  is  the  heat  rejected  to  the  cold  body,  we 
find  that 

the  entropy  lost  by  the  hot  body  ~^> 

Th 

Oh 
the  entropy  gained  by  the  working  substance  =  ^, 

■i  1 


134  THERMODYNAMICS 

the  entropy  lost  by  the  working  substance 
the  entropy  gained  by  the  cold  body 


and  that  these  are  the  only  changes  in  entropy  during  one 
complete  cycle. 

As  the  efficiency  of  this  cycle  is  less  than  the  efficiency 
of  the  ideal  cycle  considered  in  Fig.  29 

Qh-Qc>Qh-Q2, 
or  Qc<Q2, 

and  as        Ti<Th     and     7^2 > Tc  by  hypothesis, 

1  H       il  -f  2       i  C 

Also  -  ^=—,  seepage  135. 

Therefore  while  the  entropy  of  the  working  substance 
remains  unchanged  the  entropy  of  the  whole  system  now 
increases. 

Moreover  this  cycle  is  not  reversible.  It  is  impossible, 
without  aid  from  a  source  external  to  the  system  con- 
sidered, to  return  the  heat  supplied  by  the  hot  body  to  it 
by  means  of  the  work  obtained  during  the  direct  action 
of  the  cycle. 

The  above  examples  illustrate  the  fact  that  during  the 
operation  of  reversible  cycles  the  entropy  of  the  system 
remains  constant  but  that  during  the  changes  occurring 
in  an  irreversible  cycle  the  entropy  of  the  system  increases. 


ENTROPY  135 

As  all  actual  transfers  of  heat  involve  conduction  we 
may  infer  that  the  entropy  of  the  universe  is  continually 
increasing. 

The  actual  entropy  of  any  body  at  any  time  is  never 
computed  nor  need  it  be  known.  The  changes  in  entropy 
of  a  substance  due  to  a  transfer  of  heat  is  however  very 
useful.     It  is  computed  by  means  of  the  ratio 

heat  received  or  rejected  by  the  body 
the  absolute  temperature  of  the  body  during  the  transfer  of  heat' 

Let  S  represent  the  entropy  of  a  body  receiving  AQ 
heat  units  .vhile  the  absolute  temperature  of  the  body  remains 
T  then 

A5=f. 

If  the  temperature  of  the  body  changes  during  the  heat 
transfer  then 

for  during  this  transfer  of  dQ  heat  units  the  temperature 
of  the  body  remains  T. 

A  change  in  entropy  will  be  denoted  by  dS  if  the  mass 
of  the  substance  is  m  pounds  and  by  ds  if  the  mass  is  one 

pound,  so  that 

dS  =  7nds. 

The  entropy  of  a  body  may  be  used  as  a  coordinate 

to  represent  the  state  of  the  body.  The  change  in  the 
eiitropy  of  a  body  depends  only  upon  the  initial  and  the 
final  states  of  the  body  and  not  upon  the  manner  in  which 
the  change  of  state  occurred. 


136  THERMODYNAMICS 

To  demonstrate  this  Tor  an  ideal  gas  note  that 
dQ  =  mcvdT-\—:pd  V, 
and  that  by  definition 

dT     1  pdV 
so  that  db  —  mcv-jr-rj^^' 

From  this  equation  it  appears  that  the  change  in  entropy 
dS  depends  upon  T  and  pdV  in  addition  to  the  Cv  of  the 
gas.  As  pdV  represents  the  external  work  during  the 
change  of  state  considered  and  must  vary  with  the  proc- 
esses involved  it  would  seem  that  ds  must  also  vary  with 
these  processes.     But  this  is  not  so. 

As  pV  =  mRT 

thRi 
replace  p  by  — t^t-  and  note  that 

,^  dT.     R  dV 

This  equation  shows  that  the  change  in  entropy  depends 
upon  the  temperature  and  the  volume  of  the  gas  as  well 
as  upon  the  change  in  temperature  and  the  change  in 
volume  but  as  it  can  be  integrated  without  knowing  the 
relation  between  T  and  V  the  entropy  does  not  depend 
upon  the  manner  in  which  either  T  or  F  change. 

Thus  when  the  gas  attains  its  new  state  its  entropy 
reaches  a  definite  value  which  is  absolutely  independent 
of  what  may  have  happened  to  the  gas  during  its  change 
of  state. 


ENTROPY  137 

The  difference  in  this  respect  between  a  change  in  entropy 
and  say  the  heat  suppHed  during  a  change  of  state  should 
be  carefully  noted.  In  changing  from  one  state  to  another 
every  particular  process  followed  will  require  the  addition 
of  a  definite  amount  of  heat  different  in  each  case.  Thus 
Q  cannot  be  used  as  a  coordinate  to  determine  the  final 
state  of  the  gas.  With  entropy  only  one  definite  change 
in  entropy  will  be  found  during  any  change  of  state  pro- 
vided the  same  final  state  is  reached. 

Thus  instead  of  three  coordinates  to  determine  any 
particular  state  of  an  ideal  gas  as  illustrated  graphically 
in  Fig.  2  we  may  now  use  four,  i.e.,  p,  V,  T,  and  5.  This 
does  not  mean  that  all  four  must  be  used  simultaneously. 
In  fact  we  have  already  explained  that  only  two  of  the 
three  coordinates  p,  V,  and  T  are  usually  used;  instead 
of  the  three-dimensional  representation  the  two-dimensional 
representation  of  changes  of  state  on  the  />F-plane  is  always 
preferred. 

With  the  introduction  of  entropy  we  may  use  in  addition 
to  the  planes  of  pV,  pT,  and  TV,  other  planes  such  as 
pS,  VS,  TS.  Of  these  new  planes  of  projection  it  will 
be  shown  that  the  r5-plane  is  the  most  important. 

Section  XXIII 

CHANGES    IN    ENTROPY    OF    IDEAL    GASES    DURING 
REVERSIBLE    PROCESSES 

Changes  in  Entropy  during  Reversible  and  Irrevers- 
ible Processes. — It  must  be  remembered  that  changes  in 
the  entropy  of  a  substance  may  occur  even  when  no 
heat  is  supplied  to  the  substance  from  external  sources. 


138  THERMODYNAMICS 

All  heat  generated  within  the  substance  itself  will  cause 
a  change  in  the  entropy  of  the  substance.  To  illustrate, 
consider  a  mass  of  gas  flowing  thru  a  pipe.  Friction  will 
cause  some  of  the  kinetic  energy  to  be  transformed  into 
heat.  Thus  even  tho  the  gas  be  thoroly  insulated  so 
that  no  heat  may  reach  it  from  without  its  entropy  will 
nevertheless  increase.  This  is  an  example  of  an  irreversible 
adiabatic  change  of  state  during  which  the  entropy  is  not 
constant  even  tho  dQ  is  zero.  These  cases  will  be  considered 
more  thoroly  under  the  head  of  flow  of  fluids.  It  is  at 
present  sufficient  to  call  attention  to  the  fact  that  what 
follows  applies  only  to  reversible  changes  during  which 
thermal  equilibrium  exists  and  all  parts  of  the  mass  con- 
sidered are  always  at  the  same  temperature.  Neglect  of 
this  precaution  will  lead  to  serious  errors  in  the  use  of 
entropy. 

Changes  in  the  Entropy  of  an  Ideal  Gas  during  Re- 
versible Processes. — Under  these  conditions  as 

dQ  =  mc4T+jpdV 

we  have  rf5=f =«c.f +i  If. 

To  integrate  this  equation  either  p  or  V  must  be  elim- 
inated by  means  of 

pV  =  mRT. 
Eliminating  p  we  have 

dT  ,  mR  dV 
dS  =  mCv-jr-'r—j-  -yr^ 

whence         S  =  mcv  loge  T-{-m{cp—Cv)  loge  F+C 
=  maloge(rF*-i)-|-C. 


ENTROPY  139 

Assuming  that  the  entropy  changes  from  5i  to  So  while 
p,  V,  and  T  change  from  pi,  Vi,  T]_  to  p2,  V2,  To  we  have 

Si  =  mcv  log  Ti  Fi'~  ^+C 
and  ^2  =  mCv  log  T2  Vo'^'^-hC 

or  the  change  in  entropy 

S2-Si  =  mcAog(j:^j(pJ 


VoV-^ 


Exercise  137.  Show  that  the  entropy,  S,  of  an  ideal  gas 
may  be  expressed  in  the  following  forms 

S  =  nicv  log  {pV^)+Ci 
S  =  mcvlogiTV^-^)+C2 

l-k 

S  =  mcp  log  (Tp  >"  )+a. 

Exercise  138.  Show  by  means  of  the  equations  of  Exercise 
26  that  for  reversible  processes  we  may  write  the  change  in 
entropy  due  to  a  change  of  state  from  pi,  Vi,  Ti,  Si  to  p,  V,  T,  S 
in  the  following  forms 

T  V 

S-Si  =  7ncv  log-r+m{cp-Cv)  log  — , 

-11  Vi 

S—Si  =  jncp  log  - — micp—Cv)  log  — , 
J 1  Pi 

p  V 

S—Si  =  mcv  log  —  -\-mcp  log  --. 
pi  Vi 

Exercise  139.  Deduce  the  results  of  Exercise  138  from  the 
results  of  Exercise  137  and  transform  them  into  the  last  form 
given  in  the  illustrative  example  above. 


I40 


THERMODYNAMICS 


The  Temperature-Entropy  Diagram. — During  a  revers- 
ible process 

dQ=TdS. 

Therefore  the  area  under  a  line  representing  any  reversible 
process  on  the  T^-plane  such  as  i  2,  Fig.  30,  represents 
the  heat  supplied  to  the  mass  undergoing  the  change  of  state 
from  I  to  2  in  accordance  with  the  indicated  process.  If 
the  process  changes,   the  line  i  2   (Fig.  30)  must  change 


Fig.  30. 

and  the  heat  supplied  must  change  but  the  change  of 
entropy  and  position  of  point  2  remain  unchanged. 

This  graphical  representation  of  heat  supplied  on  the 
r^-plane  is  analogous  to  the  graphical  representation  of 
work  done  on  the  ^F-plane. 

The  area  enclosed  between  the  lines  representing  upon 
the  r5-plane  the  processes  of  a  reversible  cycle  represents 
the  heat  transformed  into  work  during  this  cycle. 

Reversible  Isopiestic  Processes  on  the  TS-plane. — Let 
p\,  I'l,  Ti,  51  represent  the  initial  conditions  of  one  pound 
of  gas  and  pi,  v,   T,  s  any  subsequent  condition  of  this 


ENTROPY 


141 


same  gas  after  a  reversible  isopiestic  process.  This  process 
is  represented  graphically  on  the  pv--p\a.ne  in  Fig.  31.  We 
must  now  transfer  this  representation  to  the  r.j-plane, 
Fig.  32. 


As 
and  as 

and 


dq  =  CpdT 

CndT 


ds  = 


T 


S-Si=Cy\0ge7ir. 

-t  1 


Fig.  31. 


Fig.  32. 


By  means  of  pv  =  RT,  T\  may  be  computed  and  the 
horizontal  line  on  which  point  i  lies  in  Fig.  32  may  be 
found.  The  value  of  s\  may  now  be  arbitrarily  assumed 
for  there  is  no  absolute  zero  of  entropy  such  as  exists 
for  pressure,  volume,  and  temperature.  In  practice  the 
entropy  of  the  substance  at  the  lowest  temperature  occurring 
in  any  calculation  may  conveniently  be  assumed  to  be  zero. 
This  would  place  point  i  in  Fig.  32  on  the  axis  of  T. 

T  may  also  be  computed  by  means  of  pv  =  RT.  But  5 
cannot  be  arbitrarily  assumed  after  si  is  fixed.     As 

T 


S  =  Si-\-Cp  loge 


Ti' 


142 


THERMODYNAMICS 


the  point  2  on  the  r5-plane  lies  upon  a  logarithmic  curve 
passing  thru  point  i. 

Exercise  140.  One  pound  of  air  expands  at  a  constant  pressure 
of  100  pounds  per  square  inch  absolute  from  a  volume  of  i.g 
cubic  feet  to  3  cubic  feet. 

(a)  Compute  the  change  in  entropy  of  this  air. 

(b)  Compute  the  final  entropy  of  this  air. 

The  relation  between  any  two  isopiestic  curves  on  the 

r^-plane    is    important.     The    equation    of    any    isopiestic 
curve  is 

1        ^ 

S  —  Si  =  Cp  lOge  =r. 
1  1 

By  diflferentiation  we  obtain 

ds     Cp 

T 


Fig.  33. 


Therefore   at    any  given   temperature  all  isopiestic  curves 

for  a  given  gas  have  the  same  slope,  — ,  Fig.  t^t,. 

As  an  application  of  this  fact  suppose  a  series  of  iso- 
piestics  on  the  T^-plane  are  required.  Then  one  such 
curve  may  be   accurately  plotted   and  a   templet   of   the 


ENTROPY  143 

form  indicated  in  Fig.  33  may  be  cut.  This  templet  sliding 
along  a  straight  edge  coinciding  with  the  5-axis  will  serve 
to  draw  all  the  required  isopiestics. 

Exercise  141.  Compute  the  distance,  measured  on  a  line 
parallel  to  the  5-axis,  between  two  isopiestic  curves  for  pres- 
sures pi  and  p2  and  show  that  this  distance  is  independent  of 
the  isothermal  along  which  it  is  measured. 

Exercise  142.  Isopiestics  for  o,  10,  20,  etc.  pounds  per 
square  inch  absolute  are  to  be  drawn  on  a  T^-plane.  Are  the 
distances  between  these  isopiestics  measured  along  any  given 
isothermal  equal? 

Reversible  Isometric  Processes  on  the   TS-plane. — 

It  can  be  readily  shown  that  the  change  in  entropy  during 
a  reversible  isometric  process  is 

5— ^1  =  Co  log  ^r. 
J  1 

Therefore  the  isometrics  on  the  Ts-plane  are  also  log- 
arithmic curves  but  their  slopes  are 

ds     Cv 

As  Cp>Cv  for  any  ideal  gas  isometrics  are  steeper  than 
isopiestics  on  the  T^-plane. 

Exercise  143.  Two  pounds  of  air  are  confined  in  a  receiver 
(capacity  10  cubic  feet)  at  a  temperature  of  80°  F.  The  pres- 
sure increases  to  70  pounds  per  square  inch  absolute.  Compute 
the  increase  in  the  entropy  of  this  air  during  this  change  of  state. 

Exercise  144.  Sketch  a  network  of  isopiestics  and  isometrics 
on  both  the  pv-  and  the  T^-planes.  Pay  due  attention  to  the 
slopes  and  the  relative  positions  of  the  curves  on  the  T^-plane. 


144  THERMODYNAMICS 

Reversible  Isothermal,  Adiabatic,  and  Poljrtropic  Proc- 
esses on  the  TS-plane. — Reversible  isothermals  are  evi- 
dently represented  by  straight  lines  parallel  to  the  5-axis. 
Reversible  adiabatics  for  which  dQ  =  o  and  no  heat  is 
generated  within  the  mass  considered  are  processes  for 
which  ds  =  o  so  that  the  entropy  is  constant.  These  are 
represented  by  straight  lines  parallel  to  the  T-axis. 

Exercise  145.  Show  that  the  equation  of  the  r^-curve  repre- 
senting any  reversible  polytropic  process  is 

(n-k   \         T 
\n—i    I         Tx 

From  the  last  exercise  it  follows  that  polytropics  are  also 
logarithmic  curves  on  the  T^-plane.  Note  that  isometrics, 
isopiestics,  and  the  general  polytropic  curves  all  have  the 
equation 

S-Si=ClOge;^, 

the  proper  value  of  c  to  be  introduced  in  each  case. 

The  Temperature-Entropy  Plane. — The  preceding  dis- 
cussions and  exercises  lead  to  the  results  illustrated  in  Fig. 
34.  This  figure  should  be  compared  with  the  corresponding 
diagram  for  the  /'ZJ-plane,  Fig.  4. 

Note  that  isometrics  and  isopiestics  are  straight  lines 
on  the  />i)-plane  but  logarithmic  lines  on  the  T^-plane. 
Isothermals  and  adiabatics  are  curves  on  the  ^f-plane 
but  become  straight  lines  on  the  T^-plane. 

Exercise  146.  The  state  of  an  ideal  gas  changes  from  px, 
Di,  Ti,  5i  to  pi,  V2,  T2,  S2  in  a  reversible  manner,  assume  that 
the  change  occurs  in  two  steps,  the  first  being 


ENTROPY 


145 


(a)  adiabatic  and  the  second  isometric, 

(b)  isothermal  and  the  second  isometric, 

(c)  isopiestic  and  the  second  isometric, 

(d)  isometric  and  the  second  isopiestic. 

Sketch   these   changes   on   the  pv-plane   and   note    the   relative 
amounts  of  external  work  performed. 

Exercise  147.  Sketch  the  processes  described  in  Exercise 
146  on  the  r^-plane  and  note  the  relative  amounts  of  heat  that 
must  be  supplied  to  produce  these  changes  of  state."* 


Fig.  34. 


Exercise  148.  Compute  the  changes  in  entropy  during  the 
processes  described  in  Exercise  146  and  show  that  the  total 
change  in  entropy  is  the  same  in  each  case. 

Exercise  149.  Deduce  the  relation  between  p  and  v,  v  and  T, 
and  T  and  p  for  a  reversible  adiabatic  process  by  means  of 
the  equations  in  Exercise  138. 

The  Logarithm  Temperature-Entropy  Diagram. — We 
have  found  that  polytropic  processes  are  represented  upon 
the  pv-p\a.ne  by  curves  whose  equations  are  pv**=C.  These 
curves  are  in  general  difficult  to  plot.     Their  use  does  not 


146  THERMODYNAMICS 

lead  to  simple  graphical  methods  for  the  solution  of 
problems. 

On  the  r^-plane  polytropics  are  represented  by  logarith- 
mic curves.  Also  all  polytropic  curves  having  the  same 
exponent  n  may  be  drawn  by  means  of  the  same  templet 
(see  page  142).  This  permits  graphical  calculations  to  be 
performed  more  easily  on  the  Ts-plane  than  on  the  pv- 
plane.     But  it  requires  a  templet  for  each  polytropic. 

For  all  graphical  calculations  it  is  very  desirable  to 
have  to  deal  only  with  straight  lines.  All  curves  can  be 
represented  by  straight  lines  by  properly  selecting  the 
scales  used  on  the  coordinate  axes. 

The  equation 

s—si  =  c\oge  — 
-f  1 

representing  polytropic  change  on  the  T^-plane  is  rep- 
resented by  the  logarithmic  curve  if  the  decimal  scales 
are  used  on  both  axes  (Fig.  33).  If  however  we  rewrite 
the  equation  as  follows 

S  =  C  loge  T-\-{si  —  C  loge  Ti) 

and  put  loge  r=z, 

we  obtain  s  =  cz-\-{s\  —  c\ogeTi), 

the  equation  of  a  straight  line  when  uniform  decimal  scales 
are  used  on  the  s  and  the  z  axes. 

The  scale  on  the  2-axis  would  thus  appear  as  shown  in 
Fig.  35  (a).  The  reading  of  this  scale,  in  terms  of  2  is 
easy,  in  terms  of  T  it  is  impossible  for  these  intervals  are 
non-uniform.      As  T  is  required  and  not  2  and  in  order 


ENTROPY 


147 


to  avoid  reading  z  and  then  referring  to  a  table  of  natural 
logarithms  for  the  corresponding  value  of  T  as  must  be 
done  when  scale  (a)  is  used,  the  scale  shown  in  Fig.  35  (6) 
is  used. 

Paper  ruled  horizontally  with  the       H^  h^ 

scale  shown  in  Fig.  35  (b)  and  ver- 
tically with  a  uniform  decimal  scale 
can  thus  be  used  to  plot  any  poly- 
tropic  on  the  T5-plane  as  a  straight 
line. 

Exercise  150.  If  100°  F  is  repre- 
sented by  a  point  i  inch  above  0°  F 
on  the  log  T-axis  how  far  below  the 
point  representing  0°  F  should  —  200°  F 
be  located? 

Where  would  the  absolute  zero  be 
located? 

For  diagrams  ruled  as  above  de- 
scribed and  which  may  be  used  for  the  graphical  solution 
of   problems  relating  to  air  see  the  diagrams  accompanying 
Bulletin   No.    63   of   the   Engineering    Experiment    Station 
at  the  University  of  Illinois. 


6-+ , 

5.75  ""1.75 

5-  -|-h6094 

4.48  4- 1.50 
4-  +-1-.3863-- 

■ -1.0986- - 
I 0- 


0-H — 

(a) 


3.49- -1.25 


2.72- -1.00 
_2,12;:0JjS 


(6) 


Fig.  35. 


Section  XXIV 
GAS    CYCLES   ON   THE   TS-PLANE 


As  all  reversible  processes  can  be  represented  on  the 
r5-plane  any  cycle  consisting  of  such  processes  can  also 
be  shown  on  this  plane.  The  area  included  between  the 
lines  representing  the  processes  represents  the  heat  trans- 
formed into  work  during  the  cycle. 


1 40  THERMODYNAMICS 

In  sketching  any  cycle  on  the  T^-plane  care  should  be 
taken  to  note  whether  the  temperature  rises  or  falls,  or 
whether  heat  is  supplied  to  or  rejected  by  the  gas  during 
each  process  as  it  occurs  in  the  cycle. 

Exercise  151.  (a)  Sketch  a  Carnot,  a  Stirling,  an  Ericsson, 
a  Joule,  an  Otto,  and  a  Diesel  cycle  on  the  pV-planc. 

(b)  Sketch  these  cycles  on  the  r5-plane  and  number  the 
points  so  as  to  correspond  with  the  numbers  on  the  /^F-diagrams 
drawn  in  (a). 

Exercise  152.  Compute  the  efficiency  of  a  Carnot  cycle  from 
the  areas  of  its  r5-diagram. 

Exercise  153.  Sketch  superimposed  T^-diagrams  of  a  Carnot, 
a  Stirling,  and  an  Ericsson  cycle  (the  last  two  when  regenerators 
are  used).  Assume  these  cycles  to  operate  between  the  same 
limiting  temperatures  and  assume  the  gas  to  receive  the  same 
quantity  of  heat  from  the  source  in  each  case. 

Show  by  means  of  these  diagrams  that  the  efficiencies  of  these 
cycles  are  the  same. 

Exercise  154.  Compute  the  efficiency  of  the  Otto  cycle 
from  the  areas  of  its  T^-diagram,  assume  adiabatic  expansion 
and  compression. 


VAPORS 

CHAPTER  VUI 
INTRODUCTION 

Section  XXV 
PROPERTIES    OF   VAPORS 

The  study  of  vapors  can  best  be  begun  by  an  analysis 
of  the  behavior  of  water  under  varying  conditions  of  pres- 
sure, temperature,  and  volume. 

It  has  been  determined  experimentally  that  the  pressures 
exerted  by  water-vapor  at  certain  temperatures  are  those 
given  in  the  following  table.  The  last  column  gives  the 
volume  in  cubic  feet  occupied  by  one  pound  of  liquid 
water  at  the  temperature  given  in  the  first  column. 

PROPERTIES   OF  WATER 


Temperature, 

Deg.  F. 

Pressure,  Pounds  per 
Square  Inch  Absolute. 

Specific  Volume,  Cubic 
Feet  per  Pound. 

30 

0.08 

0.01602 

120 

I 

69 

0.01620 

210 

14 

13 

0.01670 

300 

67 

00 

0.01744 

4CX3 

247 

0.0187 

500 

684 

O.CJ06 

600 

1574 

0.024 

149 


ISO 


THERMODYNAMICS 


When  the  temperature  of  and  the  pressure  on  the  water 
are  those  given  in  any  line  of  this  table  and  provision 
is  made  for  expansion  at  this  constant  pressure  then  any 
addition  of  heat  will  cause  vapor  to  form  but  no  change 
in  pressure  or  in  temperature  occurs. 

To  more  clearly  understand  the  nature  of  these  changes 
it  is  useful  to  conceive  the  water  to  be  confined  as  shown 
in  Fig.  36.  Here  the  only  pressure  exerted  upon  the  water 
is  that  due  to  the  heavy  freely  moving  piston. 


Fig.  36. 


To  fix  our  ideas  let  the  mass  of  water  be  one  pound 
and  the  weight  of  the  piston  be  such  as  to  exert  a  force  of 
say  67  pounds  per  square  inch  upon  this  water.  Then 
provided  the  temperature  of  the  water  is  less  than  300°  F 
liquid  water  alone  exists  under  the  piston;  no  water-vapor 
is  present.  When  the  temperature  reaches  300°  F  (due 
to  the  absorption  of  heat)  vapor  forms  and  more  vapor  forms 
as  more  heat  is  supplied  but  as  long  as  liquid  water  is 
present  the  temperature  remains  300°  F  and  of  course 
the  pressure  remains  67  pounds  per  square  inch  absolute. 

Consider  now  the   changes  in  volume  involved  in   the 


INTRODUCTION  151 

above  specific  case.  According  to  our  table  the  volume 
of  one  pound  of  water  at  300°  F  is  0.01744  cubic  foot. 
Altho  the  pressure  67  is  given  in  the  same  Une  of  the 
table  this  does  not  mean  that  this  must  be  the  pressure 
exerted  on  the  water  when  its  specific  volume  is  0.01744. 
Experiment  shows  that  water  is  practically  incompressible 
(0.0000469  of  one  unit  change  in  unit  volume  per  atmos- 
phere). Thus  the  volume  of  one  pound  of  water  at  300°  F 
is  0.01744  cubic  foot  under  any  pressure  likely  to  occur 
in  engineering  practice. 

Exercise  155.  (a)  What  is  the  specific  volume  of  water  at 
2,2°  F  when  under  a  pressure  of  247  pounds  per  square  inch 
absolute? 

(b)  What  will  be  the  specific  volume  of  water  under  the  same 
pressure  when  vapor  forms? 

Thus  if  in  Fig.  36  we  start  with  one  pound  of  \\-ater 
at  32°  F  its  volume  will  be  0.01602  cubic  foot  under  a 
pressure  of  67  pounds.  As  we  heat  this  water  under  constant 
pressure  its  volume  gradually  increases  to  0.01744  and 
its  temperature  to  300°  F.  Vapor  now  forms  and  the 
volume  of  the  liquid  and  vapor  rapidly  increases  until  the 
liquid  disappears  and  only  (dry  saturated)  vapor  at  300° 
F  remains.  The  volume  now  is  6.47  cubic  feet  as  set  down 
in  the  steam  tables.  If  more  heat  is  supplied  the  tem- 
perature of  the  vapor  rises  and  its  volume  increases.  Some 
volumes  and  the  corresponding  temperatures  of  the  now 
superheated  vapor,  always  under  a  pressure  of  67  pounds 
per  square  inch,  are 

6.99  cubic  feet  at  350°  F 
7 .  49  cubic  feet  at  400°  F 


152 


THERMODYNAMICS 


8.45  cubic  feet  at  500°  F 
9.38  cubic  feet  at  600°  F. 

The  above  described  changes  of  state  may  be  repre- 
sented graphically  on  the  ^z'-plane  as  shown  in  Fig.  37. 
In  this  figure  the  names  of  the  physical  states  of  the  sub- 
stance have  been  added. 

The  point  A  at  which  the  formation  of  vapor  starts, 
the  poi,nt  separating  the  liquid  from  the  liquid  and  vapor 


LiqulU   lLi(uuci  + Vapor ;\  Superb6atert 
'       "■    ^  \       Vappr 

Dry  Satiirated 
VapQr 


Fig.  37. 


condition,  lies  on  what  is  called  the  liquid  line.  The  point 
B  at  which  the  liquid  ceases  to  exist  and  at  which  the 
vapor  is  dry  saturated  lies  on  what  is  called  the  saturation 
line.  The  point  B  separates  the  saturated  condition  of  the 
vapor  from  the  superheated  condition. 

Exercise  156.  Plot,  by  means  of  the  steam  tables,  the  satura- 
tion and  the  liquid  lines  for  water  on  the  pv-p\a.ne. 
Plot  the  same  lines  on  the  pt-p\ane. 


INTRODUCTION 


153 


Section  XXVI 

THE    CHARACTERISTIC    SURFACE    OF   VAPORS 

In  order  to  illustrate  the  difference  between  the  behavior 
of  a  vapor  and  an  ideal  gas  the  general  nature  of  the  char- 
acteristic surface  of  a  vapor  plotted  in  three  dimensions 
on  axes  of  p,  v,  and  T  is  shown  in  Fig.  38.     This  figure 


I'lc;.  38. 


does  not  represent    any  particular  vapor  nor  is  it  drawn 
to  scale. 

The  portion  of  the  characteristic  surface  between  the 
liquid  line  AB  and  the  saturation  line  CD  is  cylindrical. 
Its  elements  are  parallel  to  the  z>-axis.  This  illustrates  the 
condition  noted  above,  that  at  constant  pressure  increase 


154 


THERMODYNAMICS 


in  volume  occurs  at  constant  temperature  as  long  as  some 
liquid  is  present. 

To  fix  our  ideas  let  us  follow  the  point  representing  the 
state  of  say  one  pound  of  water  while  it  is  heated  from 
32°  F  under  a  constant  pressure.  At  32°  F  the  volume 
of  this  water  will  be  less  than  LM  (or  Im)  Fig.  38,  its  state- 
point  thus  lies  in  the  horizontal  plane  thru  LM  but 
nearer  the  /?t;-plane  than  LM.  As  the  water  is  heated 
its  volume  and  its  temperature  both  increase  until  the 
temperature  reaches  the  value  corresponding  to  the  assumed 
constant  pressure  as  given  in  the  steam  tables,  then  the 
volume  of  the  water  equals  LM  and  the  state-point  is  at 
M.  Vapor  now  begins  to  form.  The  volume  of  the  water 
and  its  vapor  rapidly  increase  at  constant  temperature 
and  constant  pressure  and  the  state-point  moves  from 
M  to  N.  At  N  the  volume  of  the  dry  saturated  vapor  is 
that  given  in  the  steam  tables  for  the  given  pressure  and 
its  saturation  temperature.  The  supply  of  more  heat  now 
superheats  the  vapor  and  the  state-point  leaves  the  cyl- 
indrical surface  upon  which  it  moved  during  the  formation 
of  the  vapor.  It  now  follows  some  curve,  such  as  NP, 
lying  in  a  horizontal  plane.  Volume  and  temperature 
now  both  increase;  corresponding  values  are  given  in  the 
tables  for  superheated  steam. 

The  portion  of  the  characteristic  surface  upon  which  the 
state-point  for  the  superheated  condition  of  the  vapor 
lies  approximates  a  hyperbolic  paraboloid  (Fig.  2)  more 
and  more  closely  as  the  corresponding  condition  of  the 
vapor  recedes  more  and  more  from  the  saturated  condition. 
This  simply  means  that  very  highly  superheated  vapors  may 
be  assumed  to  obey  the  laws  of  ideal  gases. 


INTRODUCTION  1 55 

In  fact  actual  gases  under  low  temperatures  and  high 
pressures  pass  thru  the  state  of  superheated  vapor  to 
the  wet  vapor  and  the  liquid  states.  The  distinction 
between  vapors  and  gases  is  thus  one  of  degree,  highly 
superheated  vapors  are  gases  and  highly  compressed  and 
cooled  gases  are  vapors. 

The  equation  of  the  characteristic  surface  of  superheated 
vapors  is  usually  given  in  the  form  of  a  corrected  equation 
of  the  characteristic  surface  of  an  ideal  gas  in  the  form 

pv  =  RT-B, 
or  p{v-b)  =  RT-B. 

Here  B  is  the  correction  term  which  indicates  the  deviation 
of  the  superheated  vapor  from  the  corresponding  ideal 
gas,  and  h  represents  the  least  volume  (or  co-volume  as 
it  has  been  called)  of  the  substance.  This  co-volume  is 
practically  the  volume  of  the  substance  in  the  liquid  state 
at  low  temperature;  compared  to  the  volume  of  the  super- 
heated vapor  or  gas  it  is  very  small. 

As  examples,  Zeuner  for  steam  put  B  =  Cp^,  where  C 
is  a  constant,  in  the  first  equation.  Callendar  proposed 
the  equation 

P(v-b)=RT-^„ 

where  m  is  a  constant  for  all  vapors. 
Linde  proposed  for  superheated  steam 

^  rr^       /    >               ./i  =50,^00,000         „     \ 
pv  =  o.sg62T-p{i+o.ooi4p)l yg 0.0833), 

where  p  is  in  pounds  per  square  inch  absolute,  v  is  in  cubic 


156 


THERMODYNAMICS 


feet,  and  T'  =  /+459-6,  the  absolute  Fahrenheit  temperature, 
as  an  equation  expressing  experimentally  determined  facts. 

Exercise  157.  According  to  Linda's  equation  for  superheated 
steam  (o)  what  value  of  R  for  steam  corresponds  to  53.3  for 
air?  (b)  Does  steam  become  an  ideal  gas  when  sufi&ciently 
superheated? 

Section  XXVII 

ISOTHERMALS 

If  the  temperature  of  a  vapor  is  maintained  constant 
while   its   volume   is   diminished   the    changes   in   volume 


Fig.  39. 

and  pressure  are  related  as  illustrated  by  the  line  abed  in 

Fig.  39- 

In  the  superheated  region  the  pressure  rises  with  de- 
creasing volume,  ab.  Saturation  is  reached  at  b.  Further 
decrease  in  volume  does  not  cause  increase  in  pressure 
but  condensation  of  vapor.  The  vapor  becomes  wetter 
and  wetter  as  compression  contmues  until  all  vapor   has 


INTRODUCTION 


157 


been  condensed  to  liquid,  at  c.  During  the  change  of 
state  be  the  vapor  is  continually  saturated.  After  the 
vapor  has  wholly  disappeared  decrease  in  volume  is  accom- 
panied by  great  increase  in  pressure  always  at  constant 
temperature. 

Isothermals  for  higher  temperatures  have  the  general 
form  shown  in  Fig.  39.  As  already  indicated  in  Fig.  38 
the  liquid  and  the  saturation  lines  approach  each  other 
as  the  pressure  increases.  This  means  that  some  isothermal 
will  exist  for  which  the  horizontal,  straight  portion  is  re- 
placed by  a  point  of  inflection,  the  point  at  which  the 
liquid  and  the  saturation  lines  meet.  This  point  is  called 
the  critical  point  and  the  corresponding  pressure,  tem- 
perature, and  specific  volume  are  the  critical  pressure, 
the  critical  temperature,  and  the  critical  volume  of  the 
particular  gas  considered. 

Provided  the  temperature  of  any  substance  remains 
above  its  critical  temperature,  i.e.,  provided  its  state- 
point  lies  above  and  to  the  right  of  its  critical  isothermal 
(Fig.  39)  the  substance  cannot  be  liquefied. 

A  gas  may  even  be  compressed  into  a  liquid  without 
ever  condensing  any  part  of  it.  This  may  be  done  as 
follows.  Referring  to  Fig.  39,  assume  the  gas  heated 
well  above  its  critical  temperature  (.4),  compress  it  at  this 
temperature  {AB).  Under  the  conditions  represented  by 
B  the  substance  is  still  a  gas  as  its  temperature  is  still 
above  its  critical  temperature.  Next  cool  the  substance 
at  constant  pressure  until  the  temperature  falls  below 
its  critical  temperature  {BC).  The  gas  has  thus  been 
reduced  to  the  liquid  state  without  passing  thru  the  con- 
ditions of  dry  saturated  and  wet  vapor. 


158  THERMODYNAMICS 

Exercise  158.  Indicate  in  Fig.  39  the  region  in  which  the 
state-point  must  He  if  the  substance  is  to 

(a)  be  wholly  liquid, 

(b)  be  partly  liquid, 

(c)  exist  as  a  vapor, 

(d)  exist  as  a  superheated  vapor, 

(e)  exist  in  the  gaseous  state. 

The  Equation  of  Van  der  Waals. — The  continuity  of 
the  properties  of  a  substance  in  gaseous  and  liquid  states 
indicated  above  led  van  der  Waals  to  apply  his  equation 


(^p^^y,-b)=RT, 


to  the  liquid  as  well  as  the  gaseous  states  of  substances, 
altho  this  equation  was  originally  designed  to  represent 
the  deviation  of  actual  gases  from  the  ideal  conditions 
represented  by  pv  =  RT. 

In  van  der  Waals'  equation  b  represents  the  least  volume 
of  the  gas  and  a  is  a  measure  of  the  attraction  of  the 
molecules.  R,  a,  and  b  are  usually  considered  constant 
altho  a  and  b  most  likely  vary  with  both  temperature  and 
pressure. 

Van  der  Waals'  equation  is  an  attempt  to  represent 
in  one  equation  the  surfaces  of  the  saturated  and  of  the 
superheated  vapors  in  Fig.  38  together  with  the  extension 
of  the  latter  surface  into  the  gaseous  region. 

If  in  ip-\ — 2)(^~^)  — -^^j  T  is  made  constant  the  equa- 
tion of  the  ^iJ-projection  of  this  isothermal  is  obtained. 
If  in  addition  p  is  made  constant  the  intersection  of  this 
isothermal  on  the  pv-plsLue  with  the  constant  pressure  line 
on  the  same  plane  is  obtained.      Under  these  conditions 


INTRODUCTION 


159 


we  have  a  cubic  equation  in  v  and  the  three  roots  of  this 
equation   give   the   volumes   corresponding    to    the   points 
A,  B,  and  C,  Fig.  40.     Points  A  and  C  lie  upon  the  liquid 
and     the     saturation     lines 
respectively  while  B  has  only 
an   indirect    physical    inter- 
pretation.    When    the  three 
roots  of  the  cubic   equation 
are   equal    then    the    points 
A,  B,  and  C   coincide  with 
each    other    and    with     the 
critical    point    of    the    sub- 
stance.  For  any  temperature 
higher  than  the  temperature 

which  causes  A,  B,  and  C  to  coincide  the  cubic  equation 
will  have  only  one  real  root. 

Exercise  159.  Show  that  the  critical  temperature  Tc,  the 
critical  pressure  pc,  and  the  critical  volume  Vc  as  computed  from 
van  der  Waals'  equation  are 


Fig.  40. 


b    a  I  a 

27  bR  276' 


Van  der  Waals'  equation  does  not  completely  represent 
the  changes  of  state  of  a  substance  when  a  and  b  are  con- 
sidered constant  not  only  on  account  of  the  curve  between 
A  and  C,  Fig.  40,  but  also  because  no  second  jog  occurs 
in  the  isothermal  at  such  points  as  would  represent  the 
change  from  the  liquid  to  the  solid  states. 


CHAPTER  IX 

FORMATION  OF  VAPORS  AT  CONSTANT  PRESSURE 

Section  XXVIII 

DRY   SATURATED   VAPORS 

Heat  Required  to  Warm  the  Liquid. — It  is  usual  to 
assume  the  heat  of  the  substance  to  be  zero  at  32°  F. 
Referring  to  Fig.  36  assume  the  liquid  under  the  piston 
to  be  at  32°  F  and  under  any  pressure,  p  pounds  per 
square  foot.  Then  the  heat  required  to  warm  one  pound 
of  liquid  from  32°  F  to  any  temperature  /°  F,  less  than 
the  temperature  corresponding  to  the  pressure  as  given  in 
the  vapor  tables  is 

q=  )    c'dt, 

where  c'  is  the  specific  heat  of  the  liquid. 

This  specific  heat  is  the  specific  heat  at  constant  pres- 
sure and  it  varies  with  the  temperature,  but  not  appreciably 
with  the  pressure. 

The  heat  of  the  liquid  is  defined  as  the  heat  which  must 

be  added  to  one  pound  of   the  liquid  in  order  to  change 

its   temperature     from    32°    F    to   the   temperature   /'   at 

which  vapor  forms  under  the  constant  pressure  at  which 

the  liquid  is  heated. 

ft' 
Therefore  q'=  )     c'dt. 

Jli2° 

In  what  follows  a  prime  (')  affixed  to  a  letter  always 
denotes  a  condition  applying  to  liquids  and  especially  to 

160 


i"ORiLA.TION     OF   VAPORS   AT   CONSTANT   PRESSURE       l6l 

the  conditions  along  the  liquid  line.     A  double  prime  (") 
always  denotes  conditions  along  the  saturation  line. 

In  this  case  g'  is  the  heat  added  to  change  the  tem- 
perature of  the  liquid  from  32°  to  the  temperature  at  the 
liquid  line,  /',  or  which  is  the  same  thing  the  temperature 
for  the  same  pressure  at  the  saturation  line,  t". 

For  approximate  calculations  the  specific  heat  of  water 
may  be  assumed  constant  and  equal  to  unity.  Under  this 
assumption 

q'  =  t"-^2. 

For  accurate  results  the  variation  of  the  specific  heat 
of  water  must  be  recognized. 

At  40°        80°        120°       160°       200°     400°     600° 

c'  equals  1.0045    0.9970   0.9974    1.0002    1.0039    1.064    1-172 


Fig.  41. 


In  order  to  integrate  the  equation 
q'=  C  c'dt 

^32° 


either  the  variation  of  c'  must  be  expressed  as  a  function 
of  /  and  the  integral  evaluated  or  c'  must  be  plotted  with 
respect  to  /,  as  shown  in  Fig.  41,  and  the  area  under  the 
curve  must  be  determined  from  the  plot. 


l62  THERMODYNAMICS 

Exercise  i6o.  By  means  of  the  steam  tables,  find  the  dif- 

fi" 
ference  between  (/"  — 32)  and    I     c'dt  for  water  heated  at  0.505, 

5,  14.7,  100,  and  300  pounds  per  square  inch  absolute. 

The  fact  that  at  14.7  pounds  or  212°  F  the    values  of 

/"— 32  and    I     c'dt  are  alike  is  due  to  the    definition  of 

the  mean  B.t.u.  used.  It  is  the  ^\-^  part  of  the  heat  re- 
quired to  warm  one  pound  of  water  from  32  to  212°  F 
under  a  pressure  of  one  standard  atmosphere. 

The  Total  Heat  of  Dry  Saturated  Vapor.— This  is 
defined  as  the  heat  required  to  transform  at  constant  pressure 
one  pound  of  liquid,  originally  at  32°  F  and  under  the 
pressure  at  which  the  vapor  is  to  form,  into  dry  saturated 
vapor.  It  will  be  denoted  by  q".  Thus  referring  to 
Fig.  36,  we  start  with  one  pound  of  liquid  at  32°  F  under 
the  piston  which  produces  the  given  pressure  at  which  the 
vapor  is  to  be  formed.  When  q"  heat  units  have  been 
supplied  then  the  space  below  the  piston  contains  dry 
saturated  vapor. 

For  water  Regnault  found  as  the  result  of  his  experi- 
ments that 

9"=  io9i.7-i-o.305(/"-32), 

see  also  page  164. 

The  Latent  Heat  of  Vaporization. — This  is  defined 
as  the  heat  required  in  order  to  change  one  pound  of  liquid 
at  the  saturation  temperature  corresponding  to  the  con- 
stant pressure  into  dry  saturated  vapor  at  the  same  tem- 
perature. Denoting  the  latent  heat  of  vaporization  by 
r  we  have 

r  =  q"-q'. 


FORMATION   OF   VAPORS   AT   CONSTANT   PRESSURE       163 

For  water,  according  to  Regnault, 

r=  1091.7  — o.695(/"— 32). 

External  and  Internal  Latent  Heat. — During  the  ab- 
sorption of  the  latent  heat  of  vaporization  a  large  increase 
in  volume  occurs.  Therefore  external  work  must  be  done 
in  overcoming  the  constant  pressure.  The  volume  of 
one  pound  of  the  liquid  at  saturation  temperature  is  v', 
the  volume  of  one  pound  of  the  dry  saturated  vapor  at  the 
same  temperature  is  v",  the  constant  pressure  is  p,  and  the 
external  work  done  equals  p{v"—v')  foot-pounds  which 
is  equivalent  to 

^(^  E.t.u. 

Thus  of  the  r  B.t.u.  supplied  during  vaporization  only 
p{v"-v') 


J 


■=p  B.t.u. 


remain  as  increase  in   internal   energy  within   the  vapor. 
This  is  the  internal  latent  heat. 

— — — —  is  called  the  external  latent  heat,  this  repre- 
sents energy  stored  without  the  vapor. 

The  Internal  Energy. — The  internal  energy  of  a  sub- 
stance, that  is,  the  increase  of  the  internal  energy  over  its 
value  at  32°  F,  will  be  denoted  by  u. 

For  the  liquid  at  saturation  temperature  the  heat  sup- 
plied is  q'.  The  increase  in  the  specific  volume  is  v'—V32, 
where  1^32  is  the  specific  volume  of  the  liquid  at  32°  F 
which  for  water  equals  0.01602.     Therefore  if  q'  is  supplied 


164  THERMODYNAMICS 

at  a  constant  pressure  p  the  increase  in  the  internal  energy  is 

,_    ,       p{v'-V:i2) 

U-q J . 

The  last  term  is  relatively  small  and  is  usually  neglected- 
Under  these  assumptions 

u'  =  q' 

or  it  is  assumed  that  the  whole  heat  absorbed  remains 
in  the  liquid  as  increased  internal  energy  because  v'  is 
practically  equal  to  Vz2' 

Exercise  161.  Compute  the  external  work  done  during  the 
heating  of  one  pound  of  water  from  32°  F  to  saturation  tem- 
perature under  a  pressure  of  300  pounds  per  square  inch.  Com- 
pare this  with  the  heat  which  must  be  supplied. 

For  dry  saturated  vapor  the  internal  energy  is 

,/         ,/       p{v"-VZ2) 
"^=^^ ^ 

or  u"  =  u'-\-p, 

where  p  is  the  internal  latent  heat. 

The  Total  Heat  of  Dry  Saturated  Steam  as  Given  in 
Marks  and  Davis'  Steam  Tables  is  denoted  by  H  and 
is  defined  by  the  equation 

^-=^^"-1-^-0.04, 

where  H  is  expressed  in  mean  B.t.u. 

To  interpret  this  quantity  physically  it  may  be  trans- 
formed so  as  to  contain  q" . 

,,,_,,     p(v"-V32) 


As 


E=q  -\—^ 0.04. 


FORMATION  OF  VAPORS  AT  CONSTANT  PRESSURE   165 

H  therefore  is  not  equal  to  the  total  heat  as  defined  on 
page  162. 

^-j-  represents   the  heat  equivalent  of   the   work  done 

in  introducing  one  pound  of  liquid  (existing  at  32°  F  under 
zero  absolute  pressure)  under  a  piston  (Fig.  36)  producing 
a  pressure  of  p  pounds  per  square  foot  absolute. 

0.04  equals  the  heat  equivalent  of  the  work  done  by 
the  atmospheric  pressure  during  the  transfer  of  one  pound 
of  liquid  at  32°  F  and  under  zero  absolute  pressure  into 
the  cylinder  (Fig.  36)  for 


14.7X144X0.01602 

— =  0.0436. 

778  ^^ 


Therefore  H  represents  the  total  energy  (expressed  in 
heat  units)  that  must  be  supplied  in  order  to  produce  dry 
saturated  steam  at  any  pressure  from  water  originally  at 
32°  F  and  under  a  pressure  of  one  atmosphere. 

q"  represents  the  energy  which  must  be  supplied  if  the 
start  is  made  with  water  at  32°  F  and  under  the  pressure 
at  which  the  dry  saturated  steam  is  to  be  produced. 

~ 0.04,   represents  the  heat  equivalent  of  the  work 

done  by  a  feed-pump  in  introducing  the  water  at  32°  F 
into  the  boiler  in  which  it  is  to  be  heated  under  constant 
pressure.     H  is  greater  than  q"  by  this  amount. 

Exercise  162.  By  how  much  docs  //  for  water  exceed  q" 
at  300  pounds  per  square  inch  absolute? 

What  is  the  value  of  11  under  the  above  conditions? 

The    Heat    Content    of    Dry    Saturated    Vapor.— The 


1 66  THERMODYNAMICS 

heat  content  of  a  dry  saturated  vapor  is  denoted  by  i" 
and  is  defined  by  the  equation 

Its  physical  significance  becomes  more  apparent  when  we 
substitute  for  u"  its  value 


J 


thus  i"  =  q"  +  ^'' 


J   ■ 

From  this  equation  it  follows  that  the  heat  content  repre- 
sents the  heat  equivalent  of  the  total  energy  which  must 
be  expended  to  produce  the  vapor  from  liquid  initially 
at  32°  F  and  under  zero  absolute  pressure. 

p'V'lo 

At  low  pressures  — r-^  is  very  small  so  that  under   such 

conditions  we  may  assume  that 

:"      «" 
t   =q  . 

The  difference  between  H  and  i"  (i.e.,  —0.04)  is  always 
negligible  when  the  degree  of  accuracy  of  the  steam  table 
is  considered  so  that  under  all  conditions 

The  Heat  Content  of  the  Liquid  is  denoted  by  /'  and 
by  definition 

Substituting  the  value  of  u'  in  terms  of  q'  we  have 


FORMATION  OF  VAPORS  AT  CONSTANT  PRESSURE   167 

Thus    again    at    low    pressures    i'    is     practically     equal 
to  q'. 

The  total  heat  of  the  liquid,  h,  as  given  in  Marks  and 
Davis'  Steam  Tables  is  defined  by  the  equation 

h  =  u'-\-^ — 0.0436. 

Therefore  h  is  practically  equal  to  i'. 

Exercise  163.  One  pound  of  water  at  32°  F  and  under  a 
pressure  of  247  pounds  per  square  inch  absolute  is  to  be  trans- 
formed into  dry  saturated  steam  under  a  constant  pressure 
of  247  pounds. 

{a)  How  much  heat  is  required? 

{b)  By  how  much  is  the  internal  energy  of  this  water  in- 
creased during  the  above  change  of  state? 

Section  XXIX 
WET   VAPORS 

The  Quality  of  Wet  Vapors. — During  the  transformation 
of  the  liquid  into  dry  saturated  vapor  the  substance  is 
partly  liquid  and  partly  saturated  vapor.  Under  these 
conditions  the  substance  is  called  a  wet  vapor.  That  part 
of  one  pound  of  wet  vapor  which  consists  of  saturated  vapor 
is  denoted  by  x  and  this  fraction  x  is  a  measure  of  the 
quality  of  the  wet  vapor. 

At  the  temperature  of  the  wet  vapor  one  pound  of  liquid 
occupies  v'  cubic  feet  and  one  pound  of  dry  saturated  vapor 
occupies  v"  cubic  feet.  If  the  quality  of  the  wet  vapor  is 
X  its  volume  denoted  by  v  equals 

v  =  v"x+v'(i-x). 


1 68  THERMODYNAMICS 

Exercise  164.  Compute  the  volume  of  dry  saturated  vapor 
and  of  liquid  in  one  pound  of  steam  at  247  pounds  per  square 
inch  absolute,  the  quality  being  75  per  cent. 

The  Total  Heat,  Internal  Energy,  and  Heat  Content 
of  Wet  Vapor. — It  follows  from  the  definitions  already 
given  that  if  x  parts  of  one  pound  of  liquid  have  been 

evaporated 

q  =  q'-\-xr, 

u=u'-\-xp, 
and  as  i  =  M+^,  by  definition, 

we  also  have     i  =  u'-{-xp-\-—p{v"x-{-v'{i  —  x)} 

=  («'+^)+xjp+|(/'-/) 

or  i=i'-\-xr. 

Lines  of  Constant  Quality. — In  Fig.  42  are  shown  the 
liquid  and  the  saturation  lines  of  a  vapor.  Let  the  point 
A  represent  the  state-point  of  the  wet  vapor  when  its 
quality  is  x  and  its  pressure  p.     Then 

aA'^v' 

aA"  =  v" 

aA  =  v={:D"-'-c'yx-^v', 

and  as  A'A  =  aA  —  aA' 

=  {v"-v')x. 

Therefore  A' A  is  the  xth  part  of  A' A",  and  for  any  other 
pressure  we  would  have 

B'B  =  x(B'B"). 


FORMATION  OF  VAPORS  AT  CONSTANT  PRESSURE   l6g 

If  X  represents  a  constant  quality  then  the  locus  of  the 
points  A  or  B  as  determined  above  is  called  a  line  of  constant 
quality. 

Exercise  165.  On  the  plot  of  Exercise  156  (a)  draw  lines 
of  constant  quality  for  steam  whose  quality  is  (a)  50  per  cent 
and  (h)  75  per  cent. 

From  the  above  discussion  it  should  be  noted  that  a  given 
temperature  and  the  corresponding  pressure  are  not  suffi- 


FlG.  42. 

cient  to  determine  the  specific  volume  of  a  wet  vapor  (as 
was  the  case  in  an  ideal  gas).  For  wet  vapors  the  quality 
must  be  known  as  well  as  the  temperature  or  the  pressure 
(these  are  not  independent)  before  the  volume  can  be 
computed. 

Section  XXX 

SUPERHEATED    VAPORS 

The  Heat  Required  to  Superheat  Vapors  at  Constant 
Pressure. — After  the  dry  saturated  state  has  been  passed 
the  temperature  of  the  vapor  rises  with  added  heat  at 
constant  pressure.     The  specific  heat  of  superheated  vapors 


I70  THERMODYNAMICS 

varies  with  the  pressure  and  with  any  given  pressure  it 
varies  with  the  temperature. 
Let         Cp  represent  the  specific  heat  of  the  vapor, 
/",  the  saturation  temperature, 
ts,  the  temperature  of  the  superheated  vapor, 
q,  the  total  heat  of  the  superheated  vapor, 
then   q—q"  represents   the  heat   that   must   be   added   to 
one  pound  of  dry  saturated  vapor  in  order  to  increase  its 
temperature  to  ts  under  a  constant  pressure  and 

q—q"=  \,,Cpdt, 

^{Cp)m{t-t"), 

if  {cp)m  represents  the  mean  specific  heat  of  the  super- 
heated vapor  for  the  given  pressure  p  and  between  the 
temperatures  t"  and  ts.  The  range  of  temperature  {tg—t") 
is  called  the  degrees  of  superheat. 

As  both  Cp  and  {cp)m  vary  with  pressure  and  with  tem- 
perature, q—q"  is  not  easily  calculated  without  the  use 
of  vapor  tables. 

The  Internal  Energy  and  the  Heat  Content  of  Super- 
heated  Vapor. — The    internal   energy   of    a    superheated 

vapor  equals 

_       p{v—vz2) 
/       ' 


u=q— 


where  v  is  the  volume  of  the  superheated  vapor. 
The  heat  content  of  superheated  vapor  is 

pv 


FORMATION  OT  VAPORS  AT  CONSTANT  PRESSURE   171 

The  total  heat  of  steam,  h,  as  given  in  Marks  and  Davis' 
Steam  Tables  is  practically  equal  to  the  heat  content  of  the 
vapor. 

As  9  =  ^~j~ 

and  q   =t —, 

q-q"=i-i". 

Exercise  166.  (a)  From  the  steam  tables  find  the  heat  required 
to  superheat  one  pound  of  dry  saturated  steam  200°  F  at  300 
pounds  per  square  inch  absolute. 

{b)  Compute  the  mean  specific  heat  of  superheated  steam  for 
this  pressure  and  this  range  of  temperature. 

Exercise  167.  How  much  heat  must  be  added  per  pound 
of  steam,  quaHty  0.80,  in  order  to  change  it  to  steam  at  427.8° 
F  under  a  constant  pressure  of  100  pounds  per  square  inch 
absolute? 

RESUME 

The  Heat  Content  or  the  total  energy  (expressed 
in  heat  units)  required  to  change  one  pound  of  substance 
originally  at  32°  F  and  under  zero  absolute  pressure  to 
any  other  state  under  any  given  constant  pressure  is 

where  xi  is  the  change  in  the  internal  energy  (measured 
in  heat  units)  of  the  substance  due  to  the  above  change 
of  state, 

V  is  the  volume  of  one  pound  of  the  substance  in  its  final 
state,  measured  in  cubic  feet, 


172  THEKMODYNAMICS 

p  is  the  constant  pressure  under  which  the  change  of 
state  occurred,  measured  in  pounds  per  square  foot. 

The  heat  content  of  the  liquid,  i',  at  saturation  tem- 
perature is  given  in  the  vapor  tables. 

The  heat  content  of  the  dry  saturated  vapor,  i",  is 
given  in  the  vapor  tables. 

The  heat  content  of  wet  vapor,  i,  must  be  computed 
by  means  of  the  relation 

i=i'-\-xr, 

where  x  is  the  quality  of  the  vapor,  and  r  is  the  latent 
heat  of  evaporation  which  is  given  in  the  vapor  tables 
and  which  also  equals  i"—i'. 

The  heat  content  of  the  superheated  vapor,  i,  is  given 
in  the  vapor  tables. 

The    Internal   Energy    or    more    properly  the    change 

in  the  internal  energy  from  its  value  for   the  substance 

at  32°  F  and  under  zero  absolute  pressure   due   to   any 

change  of  state  is 

.     pv 
u=^-J. 

The  internal  energy  of  the  liquid,  w',  at  saturation 
temperature  is 

u  -I  —J-, 

where  i'  is  the  heat  content  of  the  liquid  at  p  the  pressure 
corresponding  to  the   temperature,   and  v'  is  the  specific 
volume  of  the  liquid  at  the  same  pressure  (or  temperature) 
as  given  in  the  vapor  tables. 
The  internal  energy  of  the  dry  saturated  vapor,  u", 


FORMATION  OF  VAPORS  AT  CONSTANT  PRESSURE   1 73 

is  sometimes  given  in  the  vapor  tables.  It  may  be  com- 
puted from 

u  -I  -— , 

where  i"  is  the  heat  content  of  the  dry  saturated  vapor 
at  p,  the  pressure  corresponding  to  the  temperature,  and 
v"  is  the  specific  volume  of  the  dry  saturated  vapor  at  the 
same  pressure  (or  temperature)  as  given  in  the  vapor  tables. 
The  internal  energy  of  wet  vapor,  u,  must  always  be 
computed  by  means  of  the  relation 

u  =  u'-\-xp, 

where  x  is  the  quality  of  the  vapor  and  p  is  the  internal 
latent  heat  which  is  sometimes  given  in  the  vapor  tables 
but  may  always  be  computed  by  means  of  the  relation 

P-r J—, 

where  r,  v" ,  and  v'  are  all  taken  from  the  vapor  tables 
for  the  pressure  (or  temperature)  for  which  p  and  u  are 
sought. 

The  internal   energy  of   superheated  vapor   must  be 
computed  from  the  relation 

.     pv 

where  i  and  v  are  the  heat  content  and  the  specific  volume 
of  the  superheated  vapor  for  the  pressure  p  at  which  u 
is  required,  all  given  in  the  vapor  tables. 


CHAPTER  X 
ENTROPY  OF  VAPORS 

Section  XXXI 
COMPUTATION    OF   THE    CHANGE   IN   ENTROPY 

The  change  in  entropy  of  a  substance  depends  only 
upon  the  initial  and  the  final  states  of  the  substance  and 
not  upon  the  process  followed  in  making  the  change. 

The  entropy  tabulated  in  the  vapor  tables  is  the  increase 
in  the  entropy  over  its  value  at  32°  F. 

By  definition  the  change  in  entropy  during  any  reversible 
process  per  pound  of  substance  is 

The  Entropy  of  the  Liquid  or  the  increase  in  the  entropy 
of  the  liquid  during  the  warming  at  a  constant  pressure 
from  32°  F  to  the  saturation  temperature  due  to  the  absorp- 
tion of  the  heat  of  the  liquid  q'  is 


*/492 


dT 

492       T 


If  c'  is  constant  and  as  ^'32  is  assumed  to  be  zero,  we 
have 

174 


ENTROPY   OF   VAPORS 


175 


Actually  c'  is  not  constant.      The    computation  of    5' 
may  be  carried  out  by  graphical  integration  of 


>/492    ^ 


492    J- 

c' 

This  form  of  the  integral  suggests  plotting  —  on  an  axis 

of  T  and  finding  the  area  under  the  curve. 
The  values  of  s'  are  given  in  the  vapor  tables. 

Exercise  168.  Plot  the  liquid  line  for  water  upon  axes  of 
absolute  temperature  and  of  entropy  by  means  of  the  steam 
tables. 

The  Entropy  of  Evaporation. — The  heat  absorbed 
during  evaporation  is  r,  the  latent  heat  of  evaporation. 
During  this  absorption  of  heat  the  temperature  remains 
constant.  Let  this  absolute  temperature  be  T,  then  the 
increase  in  entropy  during  evaporation  equals 

Y 

It  is  given  in  the  vapor  tables. 

Exercise  169.  Check  three  values  of  the  entropy  of  evapora- 
tion as  given  in  the  steam  tables. 

The  Entropy  of  Dry  Saturated  Vapor  or  the  increase 
in  entropy  of  dry  saturated  vapor  over  the  entropy  of 
one  pound  of  the  substance  at  32°  F,  denoted  by  s'\  is 
given  by 

and  may  be  found  in  the  vapor  tables. 

Exercise  170.  Plot  the  saturation  line  for  steam  by  means 
of  the  steam  tables  upon  the  plot  used  for  E.xercise  16S. 


176  THERMODYNAMICS 

The  Entropy  of  Wet  Vapors,  of  quality  x,  is  not 
given  in  the  vapor  tables.  It  must  be  computed  from 
the  relation 

5  =  5  +x— . 

Exercise  171.  Plot  lines  of  constant  quality  of  o.i,  0.5,  and 
0.8  upon  the  results  of  Exercises  168  and  170. 

The  Entropy  of  Superheated  Vapor,  s,  may  be  com- 
puted from  the  equation 


Jt"       dT         Jt' 


T 


where  T"  and  Ts  are  the  absolute  temperatures  of  saturation 
and  of  superheat  respectively.  5  is  given  in  the  vapor 
tables  for  superheated  vapors. 

Exercise  172.  Plot  the  path  of  the  state-point  representing 
the  formation  of  superheated  steam  at  a  constant  pressure  of 
(a)  40,  (i)  200  pounds  per  square  inch  absolute  starting  with 
water  heated  to  the  temperature  of  evaporation  for  the  given 
pressure  on  the  plot  already  used  for  Exercises  168,  170,  and  171. 

Section  XXXII 
THE    TEMPERATURE-ENTROPY    DIAGRAM    FOR    VAPORS 

In  this  diagram  the  state  of  the  substance  is  repre- 
sented by  its  absolute  temperature  and  by  its  entropy  per 
pound.  The  liquid  line  and  the  saturation  line  have  already 
been  plotted  in  Exercises  168  and  170.  These  are  shown 
in  Fig.  43. 


ENTROPY    OF   VAPORS 


177 


During  the  evaporation  of  a  liquid  the  state-point  rep- 
resenting the  change  of  state  on  the  Ts-plane  moves  from 
B  to  C,  Fig.  43.  The  entropy  increases  with  the  addition 
of  heat  and  the  temperature  remains  constant.  The  area 
under  BC  represents  the  heat  supplied  during  evaporation. 

If  more  heat  is  supplied  to  the  dry  saturated  vapor 
(state-point  C)  while  the  vapor  remains  under  the  same 
constant  pressure  which  existed  along  BC  the  path  of 
the   state-point   will   resemble   the   line   CD   (see   Exercise 


T 

-ye 

5/ 

A/ 

T- 

492^ 

j  1 
1  1 

s 

Fig.  43. 

172)  and  the  area  under  CD  represents  the  heat  required 
to  superheat  the  vapor  under  this  pressure. 

The  liquid  line  as  already  plotted  for  water  in  Exercise 
168  is  not  a  line  of  constant  pressure.  Variation  in  pressure 
has  however  so  little  effect  on  the  heat  necessary  to  change 
the  state  of  a  liquid  owing  to  the  very  small  increase  in 
volume  and  therefore  the  small  amount  of  external  work 
done  that  no  account  is  taken  of  the  variation  in  pressure. 
With  this  very  close  approximation  in  mind  the  liquid 
line  plotted  for  variable  pressure  may  be  regarded  as  a 
line  representing  the  change  of  state  at  constant  pressure 
for  any  pressure  during  the  heating  of  the  liquid. 


178  THERMODYNAMICS 

Thus  the  line  representing  the  change  at  constant  pressure 
starting  with  a  liquid  at  any  given  temperature  and  ending 
with  superheated  vapor  is  A  BCD,  Fig.  43. 

The  areas  under  the  lines  AB,  BC,  and  CD  extending 
down  to  the  axis  of  s,  the  line  of  absolute  zero  temperature, 
represent  the  heat  required  to  produce  the  respective  changes 
of  state,  or 

the  area  under  AB  =  q'B—q'A, 

the  area  under  BC  =  r, 

the  area  under  CD  =  qD—q"c- 

Lines  of  Constant  Quality. — These  lines  may  be  plotted 
on  the  r^-diagram  by  means  of  the  relation 

s  =  s  +x— , 

T  .  .  . 

s'  and  —  being  given  in  the  vapor  tables  s  can   be   com- 
puted for  various  temperatures  and  any  given  quality. 
Note  also  that  the  above  equation  in  the  form 

s—s  =A-f 

suggests  a  convenient  way  of  plotting  lines  of  constant 
quality. 

Lines  of  Constant  Heat  Content. — The  heat  con- 
tent of  a  wet  vapor  equals 

i=i'  -\-xr. 

Thus  for  any  given  heat  content  x  may  be  computed 
for  various  temperatures  after  i'  and  r  have  been  read 
from  the  steam  tables.     The  corresponding  entropy  need 


ENTROPY   OF   VAPORS 


179 


not  be  computed,  for  x  is  evidently  the  ratio  of  the  dis- 
tance from  B  to  the  point  sought  to  the  distance  BC, 

Fig.  43- 

In  the  region  of  superheated  vapor  the  temperature  and 
the  corresponding  entropy  of  one  pound  of  vapor  for  any 
given  heat  content  must  be  found  by  interpolation  in 
the  vapor  tables  for  superheated  vapor. 

Exercise  173.  Plot  on  the  result  of  Exercise  168  the  lines 
of  constant  heat  content  for  iioo  B.t.u.  in  the  wet  region  and 
for  1 1 50  B.t.u.  in  the  superheated  region. 

Lines  of  Constant  Volume. — For  the  wet  region  x 
the  quality  of  the  vapor  for  any  given  volume  may  be 
computed  from 

v=v"x-\-v'{i  —  x) 


or 


Thus  for  various  temperatures  find  v"  and  v'  from  the 
tables  and  compute  the  corresponding  x.  This  value  of 
X  is  sufficient  to  locate  a  point  on  the  constant  volume 
line  corresponding  to  the  temperature. 

For  the  superheated  region  the  entropy  and  the  tem- 
perature corresponding  to  any  given  volume  can  be  found 
from  the  tables  for  superheated  vapors. 

Exercise  174.  Plot  on  the  result  of  Exercise  168  the  line  of 
constant  volume  for  7.5  cubic  feet  per  pound  of  wet  and  of 
superheated  steam. 

The  following  graphical  construction  of  lines  of  con- 
stant volume  in  the  wet  region  is  interesting  because  it 
introduces    the    characteristic    surface    of    the    wet    vapor 


i8o 


THERMODYNAMICS 


referred  to  the  coordinate  axes  of  T,  s,  and  v.  It  has 
already  been  pointed  out  that  the  state  of  a  substance 
need  not  be  given  in  terms  of  p,  v,  and  T  only. 

Fig.  44  represents  the  elevation  and  the  plan  of  the 
characteristic  surface  referred  to  the  axes  of  T,  s,  and  v. 
The  line  A' A"  represents  the  T^-projection  of  the  inter- 
section of  the  surface  and  a  plane  parallel  to  the  i;5-plane. 


Fig.  44. 


We  must  find  the  nature  of  the  i)5-projection  of  this  same 
intersection  represented  on  the  figure  by  a'aa". 


As 


5  =  5'-}-x(— j 


or 


s—s  —X 


T 


it  follows  that  (.4  'A  )  =  x{A  'A  ") 


ENTROPY   or   VAPORS 


i8i 


or 

(a'b)  =  x{a'c). 

But 

v=v"x-\-v'ii- 

-x) 

or 

(v-v')^x{v"-v'). 

Therefore 

(a'b)      v-v' 
''~{a'c)"v"-v" 

whence  it  follows  that  a  (Fig.  44)  must  lie  upon  a  straight 
line  joining  a'  and  a",  or  the  projection  of  the  intersection 


Fig.  45- 


of  the  characteristic  surface  with  a  plane  parallel  to  the 
i;.s--plane  is  a  straight  line. 

As  shown  in  Fig.  45  various  rectilinear  elements  of  the 
surface  representing  the  wet  region  can  easily  be  drawn. 
Then  any  plane  parallel  to  the  T^-plane  must  contain  the 
constant    volume    line.      Three    points    on    this    line    are 


1 82  THERMODYNAMICS 

shown  projected  on  the  iJ5-trace  of  the  constant  volume 
plane. 

The  r5-projection  of  the  required  constant  volume  line 
is  obtained  as  indicated  in  Fig.  45. 

A  Complete  Temperature-Entropy  Diagram  for  any 
given  vapor  contains,  in  addition  to  the  vertical  lines 
representing  constant  entropy  and  the  horizontal  lines  rep- 
resenting constant  temperature,  lines  of  constant  quality, 
constant  degrees  of  superheat,  constant  volume,  constant 
pressure,  and  constant  heat  content  all  drawn  for  constant 
intervals  of  the  quantities  represented. 

Such  a  diagram  more  than  replaces  the  vapor  tables. 
When  sufficient  conditions  of  the  vapor  are  known  to 
determine  the  state-point  on  the  diagram  this  diagram 
not  only  allows  all  other  conditions  to  be  read  directly 
but  it  performs  graphically  all  necessary  interpolations  in 
the  superheated  region  and  all  the  calculations  involving 
quality  in  the  wet  region  which  must  ordinarily  be  per- 
formed when  the  tables  are  used. 

Moreover  the  diagram  allows  the  direct  reading  of  the 
final  conditions  due  to  various  changes  of  state  of  the 
vapor  for  which  the  various  lines  are  drawn  as  will  shortly 
be  explained. 

Temperature-entropy  diagrams  were  formerly  much  used 
but  they  have  now  been  superseded  by 

The  MoUier  Diagram. — This  diagram  contains  the 
same  line  as  the  T^-diagram  just  described  excepting  the 
constant  volume  lines  which  are  usually  omitted.  Instead 
of  plotting  these  lines  on  axes  of  temperature  and  entropy 
Mollier  devised  the  scheme  of  using  axes  of  heat  content 
and    of    entropy.     The    resulting    diagram    is    more    open 


ENTROPY    OF    VAPORS  183 

and  therefore  more  easily  and  accurately  read  than  the 
r^-diagram  in  the  region  near  the  saturation  line,  where 
such  diagrams  are  of  principal  use  to  engineers. 

A  Mollier  diagram  for  steam  accompanies  Marks  and 
Davis'  Steam  Tables  *  and  a  Mollier  diagram  for  ammonia 
has  been  drawn  by  Goodenough.f 

Exercise  175.  Plot  the  liquid  and  the  saturation  lines  on 
axes  of  heat  content  and  of  entropy.     Use  the  steam  tables. 

Exercise  176.  Compute  the  heat  content  and  the  entropy 
of  one  pound  of  wet  steam,  quality  0.50,  for  pressures  of  (a) 
200  and  (b)  one  pound  per  square  inch  absolute. 

Exercise  177.  On  the  plot  already  used  in  Exercise  175 
draw  the  constant  pressure  Unes  for  (a)  200,  and  (b)  one  pound 
per  square  inch  absolute,  also  the  constant  quality  line  for  0.50 
and  the  lines  of  constant  superheat  for  100°  and  300°  of  superheat. 

Other  Diagrams  have  been  devised.  One  using  axes 
of  specific  volume  and  "  total  heat  "  (in  this  case  the 
total  heat  H  as  defined  by  Marks  and  Davis  in  their 
Steam  Tables)  has  lately  been  drawn  by  Ellenwood  and 
published  by  Wiley.  § 

*  Marks  and  Davis,  Tables  and  Diagrams  of  Steam,  Longmans, 
Green,  &  Co.,  New  York,  1914. 

t  Goodenough,  Properties  of  Steam  and  Ammonia,  Jolin  Wiioy  & 
Sons,  Inc.,  New  York,  1915. 

■  §  Ellenwood,  Steam  Charts,  John  Wiley  and  Sons,  Inc.,  New  York, 
1914. 


CHAPTER  XI 
CHANGES   OF   STATE   OF   VAPORS 

Section  XXXIII 
WET  VAPORS 

In  this  chapter  are  to  be  considered  the  energy  changes 
occurring  during  various  changes  of  state  of  vapors.  Up 
to  the  present  only  changes  of  state  occurring  under  con- 
stant pressure  have  been  discussed. 

The  fundamental  equation  for  these  energy  changes 
during  which  thermal  equilibrium  exists,  i.e.  during  which 
the  temperature  thruout  the  vapor  is  uniform,  altho  not 
necessarily  constant,  is 

where   AQ  represents    the    heat    energy    supplied    to    the 
substance  by  some  external  source, 
Ai^,  the  change  in  internal  (intrinsic)  kinetic  energy, 
AP,  the  change  in  internal  (intrinsic)  potential  energy, 
AW,  the  external  work  performed  by  the  substance 
during  the  change  of  state  considered. 
It   is   seldom   necessary   to   separate   the   two   kinds   of 
intrinsic  energy,  so  that  we  may  put 

AU=AK-\-AP, 
184 


CHANGES   OF    STATE    OE    VAPORS  185 

where  AU  represents  the  total  internal  energy  of  the  sub- 
stance, and  the  fundamental  energy  equation  becomes 

AQ  =  AU+AW, 

or  when  one  pound  of  substance  is  considered 

Aq  =  Au-+-Aw. 

It  is  usual  to  represent  all  changes  of  state  on  either 
the  pv-  or  the  T^-plane.  These  planes  are  selected  because 
the  area  under  the  curve  representing  the  process  on  the 
/)i'-plane  represents  external  work  done  and  the  area  under 
the  r5-curve  of  the  process,  provided  the  process  is  re- 
versible, represents  the  heat  supplied  by  an  external  source. 

For  wet  vapors  we  have 

i  =  i'  -\-xr 

U=li  -f-Xp  =  t  —  —r 

U  -t  -  — 

p{v"-v') 
P  =  r-~—j  — 

v=v"x-\-v'{i  —  x) 
r 


s  =  s'-\-x^ 

Pressure  and  Temperature  Constant. — This  change  of 
state  is  represented  on  both  the  pv-  and  the  T^-planes 
in  Fig.  46. 

Let  subscripts  i  denote  initial  and  subscripts  2  final 
conditions,  then 

v\  =  v"xi-\-v'{i  —  x\) 

V2  =  v"x2-\-v'(l  —  X2) 


1 86 


THERMODYNAMICS 


or 


Vi     =1)2     —V, 


and  vi'  =  V2'  =  v'. 

Thus  the  change  in  vobame  equals 

V2  —  V]  =  (X2  —  Xi)(v"  —  V'). 


1    2 


/I      2 


1.1. 


V 

Fig.  46. 


The  heat  received  from  (or  rejected  to,  if  negative)  ai 
external  body  per  pound  of  substance  is 


Aq  =  q2-qi, 


where 

and 

so  that 
But 
and 
thus 


52  =  «2  ■ 


q\  =  ir 


pV32 
J 

PV32 
J    ' 

Aq  =  12  —  h . 
i2  =  i'-\-xir 
ix  =  i'-\-X2r, 

Aq=(x2  — Xi)r. 


CHANGES    OF    STATE    OF    VAPORS  187 

The  change  in  the  intrinsic  energy  of  one  pound  of  sub- 
stance is 

where  U2  =  u'-\-X2P     and     ii\=^ti'-\-x\p, 

so  that  Au  =  (x2  — Xi)p. 

Finally  the  external  work  performed  per  pound  equals 

Aw=—p{v2  —  v{)  =  —p{x2—xi){v"—v')  =  iS.q—L^u. 

Exercise  178.  {a)  Show  that  Mv  =  i\q  —  Mi  by  means  of  the 
above-developed  expressions. 

(6)  Find  Ag  directly  from  Fig.  46. 

Exercise  179.  The  volume  of  one  pound  of  wet  steam  at 
200  pounds  per  square  inch  absolute,  quality  0.30,  is  increased 
by  0.50  of  a  cubic  foot  at  constant  pressure.     Compute 

(a)  the  final  quahty, 

(i)  the  heat  required, 

(c)   the  change  in  intrinsic  energ}^, 

{d)  the  external  work  done. 

Volume  Constant. — The  graphical  representation  of  the 
change  of  state  of  wet  vapor  at  constant  volume  is  shown 
in  Fig.  47. 

Under  these  conditions  we  have 

V2  =  X2i'2"+  (l  —  .V2)l'2' 

and  as  vi  =  vo 

Xi{:Vx"-Vi')  +  Vi'  =  X2{v2"-V2')  +  V2'. 

Thus  the  final  quality  can  be  computed  when  the  initial 


l88  THERMODYNAMICS 

quality  and  the  initial  and  the  final  pressures  or  tempera- 
tures are  known. 

From  the  pv-ip\a,ne,  Fig.  47,  it  is  evident  that 

Aw  =  0. 

From  the   T^-plane  we  see  that  Aq  is  now  not  equal 
to  qo  —  qi- 

Exercise  180.  Show  in  Fig.  47  an  area  representing  (a)  q^, 
(b)  qi.     Is  Aq  =  q2—qi? 


Fig.  47. 

The  change  in  intrinsic  energy  is  always  equal  to 

Au  =  U2—ni 
u  =  ii'-\-xp 

AU  =  U2'  — Ui'+X2P2  — Xipi. 


and  as 


As 
we  may  also  write 


.     pv 


.        .      .      P2V-P1V 


where  the  i's  may  be  read  from  the  Mollier  diagram   and 


CHANGES   OF    STATE   OF   VAPORS  189 

the  V  from  any  diagram  which  contains  constant  volume 
curves,   or  v  may  be   computed   from    the    vapor   tables. 

Finally,  as  Aw=o 

Aq  =  Au. 

Exercise  181.  Compute  the  quality  of  one  pound  of  wet 
steam  after  the  pressure  drops  at  constant  volume  from  200 
pounds  per  square  inch  absolute,  when  the  quality  was  0.90, 
to  a  vacuum  of  28  inches  of  mercury. 

Exercise  182.  (a)  What  is  the  volume  of  the  steam  during 
the  change  of  state  described  in  Exercise  181? 

(b)  At  what  temperature  and  at  what  pressure  would  this 
steam  become  dry  saturated,  its  volume  remaining  constant? 

(c)  Solve  (a)  and  (b)  by  means  of  the  steam  diagrams. 
Exercise    183.  Compute   the   heat   absorbed   by   the   steam 

during  the  change  of  state  indicated  in  Exercise  181. 

Exercise  184.  (a)  Compute  the  entropy  of  one  pound  of 
steam  at  90  poirnds  per  square  inch  absolute,  the  volume  being 
4.5  cubic  feet. 

(b)  Solve  (a)  by  means  of  the  steam  diagrams. 

Entropy   Constant,   Reversible   Adiabatic   Change. — 

By  definition  A^  =  o, 

thus  Au=—Aw. 

The  process  is  represented  in  Fig.  48.  To  compute  the 
final  quality  use  is  made  of  the  fact  that  the  entropy 
remains  constant.     Thus 

from  which  X2  may  be  found. 


igo  THERMODYNAMICS 

Knowing  xo  the  final  volume  is  found  from 

v-z  —  V'/'x-z-hv-Zii  —  X'^). 
The  change  in  intrinsic  energy  is  always 

Au  =  l{2  —  Ui  =  {u2'-\-X2P2)—  (Wl'  +  ^lPl) 

and  the  external  work  done  must  equal 

Aw  =  —  Am. 
V 


|Ag=o 

I 
J 


Fig.  48. 

The  Mollier  diagram  does  not  give  the  values  of   the 
intrinsic  energy  but  as 

pv 


u  =  t— 


J 


Au=i2  —  ii  — 


P2V2—P1V1 
J 


in  which  all   quantities  are  either  given  or  may  be  read 
from  the  vapor  diagrams. 

Exercise  185.  Compute  the  quality  of  steam  after  a  reversible 
adiabatic  expansion 

(a)  from  dry  saturated  steam  at  245  pounds  per  square  inch 
absolute  to  15  pounds  per  square  inch  absolute,  ::  :* 


CHANGES    OF    STATE    OF    VAPORS  IQI 

(b)  from  245  pounds  per  square  inch  absolute  with  a  quality 
of  0.20  to  15  pounds  per  square  inch  absolute. 

(c)  Check  the  results  by  means  of  the  Mollier  diagram. 

As  the  results  of  the  above  exercise  show,  the  quality 
of  low-quality  steam  is  increased  and  the  quality  of  high- 
quality  steam  is  decreased  by  adiabatic  expansion.  The 
r^-diagram  for  steam  illustrates  these  facts.  These  con- 
ditions do  not  hold  for  all  sub- 
stances. For  ether  the  liquid  and 
the  saturation  lines  both  slope 
in  the  same  direction,  as  shown 
in  Fig.  49. 

Exercise  186.  Show  that  dry 
saturated  steam  condenses  during 
adiabatic    expansion    and    does   not  Fig.  40. 

condense     during     adiabatic      com- 
pression.    How    does  ether    behave    under    the    same  condi- 
tions? 

Exercise  187.  Compute  the  volume  of  one  pound  of  wet 
steam  before  and  after  reversible  adiabatic  expansion  from 
350  pounds  per  square  inch  absolute  and  a  quality  of  0.90  to 
50  pounds  per  square  inch  absolute. 

Check  by  means  of  the  steam  diagrams. 

Exercise  188.  Compute  the  work  performed  during  the 
reversible  adiabatic  expansion  of  10  cubic  feet  of  steam,  quality 
0.90,  from  a  pressure  of  80  pounds  per  square  inch  absolute  to 
an  absolute  pressure  of  10  inches  of  mercury. 

Exercise  189.  Solve  Exercise  188  by  means  ci  the  steam 
diagrams. 

The  pv-curve  representing  the  reversible  adiabatic  of 
wet  steam  is  of  the  approximate  form  shown  in  Fig.  48. 


192  THERMODYNAMICS 

For  an  initial  quality,  xi,  lying  between  0.70  and  i.oo 
the  equation  of  this  curve  may  be  expressed  by 

pv"*=a,  constant. 

Rankine  assumed  m  =  —, 
9 

Zeuner  placed        m=  i. 035+0. ixi, 

and  calculations  based  upon  Marks  and  Davis'  Steam 
Tables  seem  to  show  that 

m=  1.059— 0.0003 1 5/)i  +  (o.o7o6+o.ooo376/'i).vi. 

Thus  m  may  vary  between  i.io  and  1.13  in  problems 
usually  arising  in  engineering. 

As  the  area  under  any  curve  having  the  equation 
pi^=2i  constant,  where  w  is  constant,  equals 

plV\  —  p2'V2 

m—i 

this  expression  affords  another  method  of  computing  the 
work  done  during  a  reversible  adiabatic  change  of  state  of 
wet  steam.  This  method  is  not  as  accurate  as  the  method 
based  on  the  vapor  tables  on  account  of  the  questionable 
value  of  m  and  even  of  the  equation  pv^  =  a.  constant. 

Exercise  190.  Using  Zeuner 's  value  of  m,  solve  Exercise  188. 

Quality  Constant. — A  change  in  state  at  constant 
quality  is  illustrated  in  Fig.  50.  Under  these  conditions 
the  final  volume  is  evidently 

V2  =  V2"x-\-V2{l  —  X), 

where  V2"  and  V2'  correspond  to  the  final  temperature 
or  pressure  and  x  is  the  constant  quality. 


CHANGES   OF    STATE    OF   VAPORS 

The  final  entropy  is 


193 


52  =  ^2'+^; 


T 


In  all  changes  of  state  heretofore  discussed  either  one 
or  all  of  the  quantities  Ag,  Aw,  or  Aw  could  be  computed 
by  means  of  the  vapor  tables.  This  is  not  possible  for 
all  changes  of  state.  The  case  under  consideration  is  an 
illustration. 

Here  Aq  is  not  equal  to  either  qo  —  qi  or  io  —  ii. 


Fig.  50. 


The  change  in  intrinsic  energy  is  however  equal  to 
Am=M2— "1 

=  (H2'  —  Xp2)—{ui'  —  Xpi) 


_      .         p2V2\        (.         P\V{ 


The  external  work  done  cannot  be  computed  from  the 
relation  ^q  =  Au-\-\w  for  A^  is  not  known. 

In  order  to  find  Aw  the  equation  of  the  pv-cnrxe  represent- 
ing the  process  must  be  found.     It  has  been  found  empirically 


194  THERMODYNAMICS 

that  the  equation  of  the  saturation  line  on  the  pv-plane 
may  be  very  closely  represented  by  the  equation 


Pit 

t"y^'LL  constant 

where 

5=  1.063. 

As 

v  =  v"x-i-v'{i  —  x) 

X 

and  the  equation  of  a  constant  quality  line  is 

v~v'(i  —  x) ,  ^     ^ 

=  a  constant. 

If  the  pressures  are  relatively  low  and   the  quality  high 

the  term  v'{i  —  x)  is  small  compared  to  v  and  this  equation 

approximates 

pf 

~=3i  constant. 

X 

By  means  of  this  equation  an  approximate  value  of  Aw 
may  be  computed  from  the  area  under  the  curve  and  thus 
a  value  of  Aq  may  be  found. 

Exercise  191.  One  pound  of  steam  expands  with  a  constant 
quality  from  a  volume  of  2  cubic  feet  and  a  pressure  of  200 
pounds  per  square  inch  absolute  to  18  pounds  per  square  inch 
absolute. 

(a)  Find  the  equation  of  its  pv-cxirve. 

(b)  Compute  the  external  work  performed. 

(c)  How  much  heat  was  supplied  during  this  change  of  state? 


CHANGES   OF   STATE   OF   VAPORS 


195 


Section  XXXIV 
SUPERHEATED    VAPORS 

Pressure  Constant. — The  change  in  specific  volume 
and  in  entropy  for  a  given  change  in  temperature  during 
a  change  of  superheated  vapor  at  constant  pressure,  illus- 
trated in  Fig.  51,  may  be  obtained  from  the  vapor  tables 
by  interpolation  or  may  be  read  directly  from  the  vapor 
diagrams. 


Fig.  51. 

As  may  be  seen  from  Fig.  51,  the  heat  supplied  during 
this  change  of  state  is 

Aq=q2—qi  =  72  —  h- 

The  change  in  intrinsic  energy  is  always 


pl'y 

T 


and  the  external  work  done 

p{V2  —  Vi) 

Aw=— — = — -. 


-{ii- 


J 


196 


THERMODYNAMICS 


Exercise  192.  (a)  How  much  heat  must  be  suppHed  to 
change  one  pound  of  dry  saturated  steam  to  superheated  steam 
at  a  temperature  of  600°  F  under  a  constant  pressure  of  200 
pounds  per  square  inch  absolute? 

(5)  What  is  the  mean  specific  heat  of  the  steam  under  the  above 
conditions? 

Exercise  193.  Compute  the  increase  in  the  intrinsic  energy 
during  the  change  of  state  described  in  Exercise  192. 

Temperature  Constant. — If  the  temperature  and  the 
initial  and  the  final  pressures  are  known,  the  initial  and 


V/      ^ 

V/////A     s 


Fig.  52. 


the    final  volumes  and  entropies  of  one  pound  of   vap- r 
may  be  found  by  means  of  the  vapor  tables  or  the  vapo' 
diagrams. 
As  Fig.  52  shows 

Aq^qo  —  qi. 

Exercise  194.  Show  172  and  qi  on  Fig.  52. 
The  heat  supplied  is,  from  Fig.  52, 

Aq=T(s2  —  si). 

The    intrinsic    energy    is    not    constant    even    tho  the 


CHANGES   OF   STATE    OF   VAPORS 


197 


temperature  remains  constant  for  the  superheated    vapor 
is  not  an  ideal  gas.     Here  as  always 


T 


=  I  ^2  — 


pivi\ 


To  compute  the  external  work  done  we  have 
Aw^  Aq  —  All. 

Exercise  195.  One  pound  of  steam  expands  at  constant  tem- 
perature from  300  pounds  per  square  inch  absolute  and  50^ 
superheat  to  40  pounds  per  square  inch  absolute.     Compute 

(o)  the  final  degrees  of  superheat. 

(b)  the  increase  in  volume, 

(c)  the  heat  supplied, 

{d)  the  change  in  intrinsic  energy, 
(e)   the  external  work  performed. 


Fig.  53. 

Volume    Constant. — In    this   case    Au'  =  o  and  Aq  =  Au. 
But  be  careful  to  note  that  as  Fig.  53  shows 

^q7^q2  —  q\9^  12— h- 
Again  Au=U2  —  ui—  ( 12  — ^y-]—  ( 'i~^)- 


198 


THERMODYNAMICS 


Exercise  196.  During  a  reduction  in  the  pressure  on  one  pound 
of  steam  from  290  pounds  per  square  inch  absolute  and  140° 
superheat  to  235  pounds  per  square  inch  absolute,  the  volume 
remained  constant.     Compute  the  heat  supplied. 

Entropy    Constant,    Reversible    Adiabatic    Change. — 

The  changes  in  temperature,  pressure,  and  volume  of 
superheated  vapor  during  a  reversible  adiabatic  change 
of  state,  Fig.  54,  can   be  found  from  the  vapor  tables   by 


Fig.  54. 

interpolation  or  can  be  read  directly  from  the  diagrams  if 
it  be  remembered  that  the  entropy  remains  constant. 
Under  these  conditions  we  have 

Aw  =  —  Aw, 
and  AM=i<2  — wi 

P'lV-lX       ( ■       p\V\ 


=  \12-- 


Exercise  197.  One  pound  of  steam  under  200  pounds  per 
square  inch  absolute,  superheated  160°  F,  expands  adiabatically 
to  100  pounds  per  square  inch  absolute.     Compute 

(a)  the  final  degrees  of  superheat, 

{b)  the  work  performed  during  expansion. 

(c)   Check  the  results  by  means  of  the  steam  diagrams. 


CHANGES   OF    STATE   OF   VAPORS  1 99 

The  equation  of  the  /)y-curve  representing  reversible 
adiabatic  change  of  state  of  superheated  steam  has  been 
empirically  determined  to  be 

^^,i.3i_.g  constant. 

The  external  work  performed  may  thus  also  be  computed 
by  means  of 

_plVl  —  p2V2 

altho    this    method    is    not    as    accurate  as  the  method 
based  upon  the  vapor  tables. 

Exercise  198.  Solve  Exercise  197  by  means  of  the  above 
formula. 

Exercise  199.  Steam  initially  under  160  pounds  per  square 
inch  absolute  and  superheated  100°  F  expands  with  constant 
entropy  to  0.50  pound  per  square  inch  absolute.     Compute    - 

(a)  the  final  condition  of  the  steam, 

(b)  the  external  work  performed. 

(c)  Check  these  results  by  means  of  the  steam  diagrams. 
Exercise  200.  Assuming  the  equation  of  the  pv-curve  of  the 

saturation  line  of  steam  to  be 

p{v")^-^^^  =  ii  constant 

and  that  of  the  adiabatic  expansion  line  of  superheated  steam 
to  be 

pv^'^^  =  a,  constant, 

find  the  pressure  at  which  the  steam  in  Exercise  199  becomes 
dry  saturated. 

Check  this  result  by  means  of  the  steam  diagrams. 


CHAPTER  XII 

VAPOR   CYCLES 

Section  XXXV 

THE   CARNOT   CYCLE 

The  Carnot  cycle  for  ideal  gases  has  been  described 
on  page  86.  This  cycle  always  consists  of  two  isothermal 
and  two  adiabatic  processes.  For  vapors  this  cycle  can 
best  be  represented  on  the  r5-plane  as  shown  in  Fig.  55. 

Here  the  vapor  thruout  the 
various  processes  remains  wet 
vapor. 

The  whole  cycle  may  be 
conceived  to  occur  while  the 
vapor  remains  in  one  cylin- 
der. Starting  with  the  state 
represented  by  the  point  i 
the  first  process  consists  of 
the  reversible  adiabatic  compression,  i  2,  during  which 
the  quality  decreases  with  the  volume  while  the  pressure 
increases  with  the  temperature.  During  this  process  no 
heat  is  supplied  to  or  rejected  by  the  vapor  but  work  must 
be  done  on  the  vapor.  During  the  next  process  2  3  heat 
is  supplied  at  the  constant  temperature  of  the  hot  body,  Tb- 
If  the  vapor  remains  wet  the  pressure  must  remain  con- 
stant while  the  volume  increases.  Then  follows  a  reversible 
adiabatic  expansion   3  4  during  which  the  quality  of  the 


Fig.  55. 


VAPOR   CYCLES  20I 

vapor  diminishes  as  the  temperature  drops  to  the  tem- 
perature of  the  cold  body,  Tc.     During  both  the  processes 

2  3  and  3  4  external  work  is  done  by  the  vapor.  Finally 
the  wet  vapor  decreases  in  volume  while  in  contact  with 
the  cold  body  at  a  constant  temperature  Tc  (and  necessarily 
constant  pressure)  until  the  quality  returns  to  the  initial 
quality  at  i. 

Actually  the  processes  above  described  would  hot  be  per- 
formed in  one  cylinder  for  this  would  require  the  heating 
and  the  cooling  of  the  cylinder  between  the  extreme  tem- 
peratures and  this  would  entail  a  waste  of  heat.  The 
evaporation  2  3  would  be  performed  in  a  boiler,  the  con- 
densation 4  I  in  a  condenser,  and  the  adiabatic  expansion 

3  4  in  one  cylinder,  the  working 
cylinder,  while  the  adiabatic 
compression  i  2  would  be  per- 
formed in  another  cylinder 
that  of  the  feed-pump. 

The  /)i>-diagram  of  this  cycle 
would  appear  as  shown  in 
Fig.  56.  Here  the  wet  vapor 
would     leave     the    feed-pump 

with  a  volume  b  2  and  return  from  the  boiler  to  the 
working  cylinder  with  the  volume  b  3.  After  adiabatic 
expansion  3  4  the  wet  vapor  would  leave  the  engine  cylinder 
with  a  volume  4  a  and  after  passing  thru  the  condenser 
would  return  to  the  feed-pump  with  a  volume  a  i. 

No  heat  engine  has  been  built  to  operate  upon  this 
cycle  for  it  would  require  a  stoppLng  of  condensation 
at  a  definite  quality  while  in  practice  complete  conden- 
sation of  the  vapor  is  by  far  a  simpler  operation. 


202  THERMODYNAMICS 

However  as  has  already  been  shown  the  Carnot  cycle 
gives  an  efficiency  which  cannot  be  surpassed  by  any  other 
cycle  and  therefore  sets  a  standard  which  it  is  desirable 
to  approach  as  closely  as  possible,  see  page  122. 

Exercise  201.  Sketch  on  both  the  pv-  and  the  T^-planes 
a  Carnot  cycle  in  which  the  lowest  quality  of  the  vapor  is  zero 
and  the  vapor  during  its  adiabatic  expansion  (o)  changes  from 
the  superheated  to  the  wet  state,  {b)  remains  superheated,  for 
both  water  and  ether. 

Exercise  202.  Show  from  the  T^-diagram  that  the  efficiency 
of  the  Carnot  cycle  is  under  all  conditions 

Th-Tc 


Th     ' 

where  Th  is  the  absolute  temperature  of  the  hot  body  and  Tc 
is  the  absolute  temperature  of  the  cold  body. 

Exercise  203.  A  Carnot  cycle  operates  with  steam  between 

(a)   150  and  15  pounds  per  square  inch  absolute, 

{b)   1 50  and  i  pound  per  square  inch  absolute, 

(c)    200  and  I  pound  per  square  inch  absolute. 

Compute  the  efficiencies. 

The  Efficiency,  the  Mean  Effective  Pressure,  the 
Heat  Consumption,  and  the  Mass  of  Substance,  the  last 
two  per  hour  per  indicated  horse-power,  may  readily  be 
computed  for  any  cycle  in  the  following  manner. 

The  thermal  efficiency  of  a  cycle  is 

^~ — 7. — ' 

where  qn  is  the  heat  received  from  the  hot  body  and  qc 
is  the  heat  rejected  to  the  cold  body  per  pound  of  the 
working  substance. 


VAPOR   CYCLES 


203 


As  the  work  performed  per  pound  of  the  working  sub- 
stance passing  thru  the  cycle  is  J{qH—qc),  the  mean 
effective  pressure  may  be  found  by  dividing  this  work  by 
the  largest  volume  of  one  pound  of  the  substance  at  the 
lowest  pressure  of  the  cycle. 

One  horse-power  equals  33,000  foot-pounds  per  minute 
and  this  is  equivalent  to  2545  B.t.u.  per  hour.  Even  under 
ideal  conditions  a  working  substance  passing  thru  a 
cycle  can  only  transform  a  certain  fraction  rj  of  the  heat 
supplied  to  it  by  the  source  into  mechanical  energy  (see 
page  128).  Therefore  the  heat  which  must  be  supplied 
per  hour  in  order  to  yield  one  indicated  horse-power  under 
the  conditions  under  which  77  is  computed  or  the  heat 
consumption  equals 

-^  B.t.u.  per  i.h.p.  hour. 

The  mass  of  the  substance  which  must  pass  thru 
the  cycle  per  hour  per  indicated  horse-power,  measured 
in  pounds,  is  evidently  equal  to  the  heat  supplied  per  i.h.p. 
hour  divided  by  the  heat  supplied  per  pound  of  substance. 

Exercise  204.  Compute  the  work  per  pound  of  steam  and 
the  steam  consumption  in  pounds  per  indicated  horse-power 
hour  when  steam  is  used  in  a  Carnot  cycle  between  150  and  3 
pounds  per  square  inch  absolute  with  a  quahty  varying  between 
o  and  I. 

Section  XXXVI 

THE   RANKINE    CYCLE 

It  has  not  been  found  practicable  to  transform  heat 
into  mechanical  energy  by  means  of  a  vapor  carried  thru 
a  Carnot  cycle  of  changes  of  state. 


204 


THERMODYNAMICS 


The  cycle  which  approximates  most  closely  the  cycle  at 
present  used  in  heat  motors  employing  vapors  as  working 
substances  is  the  Rankine  cycle,  illustrated  in  Fig.  57. 
During  this  cycle  the  heat  is  received  and  rejected  by  the 
substance  while  it  is  under  constant  pressure.  This  cycle 
further  differs  from  the  Carnot  cycle  in  so  far  that  the  con- 
densation of  the  vapor  is  continued  until  the  vapor  is  com- 
pletely liquefied,  process  4  i.  Fig.  57. 

The  liquid  is  then  heated  under  a  constant  pressure 
corresponding  to  the  highest  pressure  in  the  cycle,  process 


Fig.  57. 


I  2,  then  vaporized  in  part,  or  wholly,  or  even  possibly 
superheated,  all  under  the  highest  constant  pressure  in 
the  cycle,  process  2  3,  then  expanded  under  reversibly 
adiabatic  conditions  3  4,  and  finally  cooled  and  condensed 
to  the  liquid  state  under  the  lowest  pressure  existing  in 
the  cycle. 

It  will  be  noted  that  the  process  i  2  as  shown  in  Fig. 
57  is  not  a  constant  pressure  process.  Actually  the  pres- 
sure suddenly  increases  (as  shown  by  the  dotted  line) 
due  to  the  action  of  the  feed-pump  and  the  heat  is  then 
supplied  to  the  liquid  under  a  constant  pressure  until  vapor 


VAPOR   CYCLES  205 

forms  at  2.  The  line  i  2  is  however  so  close  to  the  p-SLxis 
and  so  nearly  vertical  (due  to  the  small  volume  and  the  small 
change  in  the  volume  of  liquid)  that  this  distinction  need 
not  be  made  as  it  does  not  appreciably  affect  the  calcu- 
lations. 

During  the  heating  of  the  liquid,  process  i  2,  Fig.  57, 
the  temperature  of  the  liquid  is  below  the  temperature  of 
the  hot  body,  therefore  the  criterion  for  best  efficiency, 
page  100,  is  not  fulfilled  and  the  efficiency  of  the  Rankine 
cycle  must  be  less  than  the  efficiency  of  the  Carnot  cycle. 
For  the  same  reason  the  Rankine  cycle  is  irreversible  in 
the  thermodynamic  sense. 

The  pv-disLgram  of  the  Rankine  cycle  must  not  be  con- 
fused with  the  indicator  diagram  showing  the  conditions 
existing  in  the  cylinder  of  an  engine  operating  under  this 
cycle.  As  actually  carried  out  in  practice  the  processes 
shown  in  Fig.  57  do  not  occur  while  the  vapor  is  in  the 
same  cylinder.  The  processes  i  2  and  2  3  are  performed 
in  the  feed-pump,  the  boiler,  and  the  superheater,  the 
process  3  4  occurs  in  the  engine  cylinder,  and  the  process 
4  I  in  the  condenser. 

Fig.  58  shows  the  action  of  the  engine  cylinder.  Here 
J  c  is  the  ideal  reversible  adiabatic  expansion  corresponding 
to  3  4,  Fig.  57.  Only  part  of  the  ex-panded  vapor  is  now 
displaced  from  the  cylinder.  The  volume  g  d  remains  in 
the  cylinder  and  is  compressed  adiabatically  to  the  volume 
fa.  The  quality  of  the  vapor  at  d  is  the  same  as  the  quality 
at  c  (or  4,  Fig.  57)  while  at  a  the  state  is  the  same  as 
the  state  of  the  vapor  about  to  be  admitted  to  the  cyl- 
inder, i.e.,  the  state  at  b  (or  3,  Fig.  57).  Thus  vapor 
whose  volume  is  d  c  (Fig.   58)   and  whose  quality  is  the 


2o6 


THERMODYNAMICS 


same  as  the  quality  at  4  (Fig.  57)  has  been  turned  into  the 
condenser,  then  as  a  liquid  after  condensation  into  the 
boiler,  and  finally  is  returned  to  the  cylinder  with  a  volume 
a  b  (Fig.  58)  and  superheated  as  indicated  at  3  (Fig.  57). 


Fig.  58. 


Exercise  205.  Show  from  Fig.  57  that  the  net  work  performed 
by  one  pound  of  vapor  during  a  Rankine  cycle  is  very  nearly 
J{i3—ii)  foot-pounds,  that  the  hot  body  supplies  very  nearly 
ii—ii  B.t.u.  during  the  cycle,  and  that  therefore  the  efficiency  of 
the  cycle  is 

V  =  -. r. 

ts  —  ti 

Exercise  206.  An  ideal  engine  operates  under  a  Rankine 
cycle  between  160  and  i  pound  per  square  inch  absolute  and 
the  cylinder  is  supplied  with  dry  saturated  steam.     Compute 

(a)  the  thermodynamic  efficiency, 

{b)  the  work  obtained  per  pound  of  steam, 

(c)   the  steam  consumption. 

Exercise  207.  Compare  the  results  of  Exercise  206  with 
the  corresponding  values  for  an  engine  operating  under  a  Carnot 
cycle  during  which  the  steam  is  returned  to  the  boiler  in  the 
liquid  form  and  suppUed  to  the  cyUnder  as  dry  saturated  steam. 


VAPOR   CYCLES 


207 


Exercise  208.  Compare  the  results  of  Exercise  206  with  the 
results  obtained  when  the  steam  in  the  Rankine  cycle  is  supplied 
to  the  cylinder  superheated  200°  F  instead  of  dry  saturated. 

Exercise  209.  Sketch  on  both  the  pv-  and  the  T^-planes 
a  Rankine  cycle  during  whose  adiabatic  expansion  the  steam 
remains  superheated. 

The  Rankine  Cycle  with  Incomplete  Expansion. — The 

large  specific  volume  of  steam  at  low  pressures  after  expan- 
sion makes  it  impracticable  to  expand  steam  in  the  cylinders 
of  a  reciprocating  engine  down  to  a  very  low  back  pressure 


Fig.  59. 


as  shown  in  Figs.  57  and  58.  The  large  cylinders  required 
for  complete  expansion  cause  large  friction  and  radiation 
losses  which  more  than  neutralize  the  gain  even  if  the 
size,  cost,  and  weight  of  the  required  engines  were  not  pro- 
hibitive. 

In  reciprocating  engines  the  steam  is  released  at  some 
pressure  higher  than  the  back  pressure  (5,  Fig.  59)  and 
the  toe  of  the  indicator  card  is  cut  ofi".  The  steam  changes 
its  state  at  constant  volume  during  the  added  process  5  6. 

In  the  steam  turbine  these  disadvantages  accompanying 
complete    expansion  do  not  exist,  in  fact  the  losses  due  to 


2o8 


THERMODYNAMICS 


leakage  past  the  blades  and  due  to  friction  (windage) 
diminish  with  increasing  specific  volume  of  the  steam  so 
that  the  steam  turbine  is  particularly  fitted  to  utilize  com- 
plete expansion  to  very  low  back  pressures. 

Steam  can  therefore  be  employed  most  advantageously 
in  cylinders  of  reciprocating  engines  at  high  pressures  and 
in  turbines  at  low  pressures.  For  this  reason  the  exhaust 
of  high-pressure  reciprocating  engines  can  be  used  in  low- 
pressure   turbines   with   higher   efficiencies   than   could   be 


Fig.  6o. 

obtained  with  either  an  all  reciprocating  or  an  all  turbine 
installation. 

The  r^-diagram  of  the  Rankine  cycle  with  incomplete 
expansion  (Fig.  60)  shows  the  loss  in  efficiency  due  to  the 
drop  in  pressure  at  constant  volume.  Note  that  the  work 
performed  with  the  same  heat  supply  is  diminished  by  the 
area  546  due  to  incomplete  expansion.  The  extreme 
case  in  which  the  steam  is  used  non-expansivelyis  shown  by 

the  cycle  i  2  3  7  i. 

Exercise  210.  Compare  the  results  of  Exercise  206  with 
the  results  obtained  under  the  same  conditions  with  the  excep- 
tion that  the  expansion  is  continued  to  only  21  pounds  per 
square  inch  absolute. 


VAPOR    CYCLES  SOQ 

Section  XXXVII 
ENGINE   EFFICIENCIES 

The  thermal  efnciencies  computed  in  the  exercises  of  the 
previous  sections  do  not  give  the  complete  facts  concerning 
the  operation  of  an  actual  engine.  These  efficiencies  were 
computed  under  the  assumptions  that  the  engine  parts 
were  non-conducting,  that  the  expansions  were  reversibly 
adiabatic  (occurred  with  unchanging  entropy),  and  that 
no  losses  due  to  throttling  or  wiredrawing  occurred. 

As  the  temperature  of  the  steam  varies  during  expansion 
the  temperature  of  the  cylinder  must  also  vary.  The  high- 
pressure  steam  entering  the  cold  cylinder  must  heat  it. 
This  causes  a  condensation  of  some  of  the  incoming  steam 
and  not  only  reduces  the  quality  of  this  steam  below  the 
quality  in  the  steam-main  but  necessitates  a  greater  supply 
of  steam  per  stroke  for  the  actual  cylinder  than  would  be 
the  case  in  an  ideal  non-conducting  cylinder.  Some  of 
this  condensed  steam  is  re-evaporated  at  the  end  of  the 
stroke  after  the  steam  in  the  cylinder  has  been  cooled 
by  expansion. 

Several  methods  may  be  used  to  overcome  at  least 
partly  this  cylinder  condensation  which  greatly  lowers  the 
thermal  efficiency  of  the  engine.  One  of  these  is  the  use 
of  superheated  steam.  As  Exercise  208  shows,  but  little 
is  gained  in  efficiency  under  ideal  conditions  by  the  use 
of  superheated  steam.  Practically  however  a  greater  gain 
is  produced  due  to  the  superheated  steam  losing  its  super- 
heat in  heating  the  cylinder  and  yet  allowing  ex-pansion 
to  begin  with  practically  dry  saturated  steam. 


2IO 


THERMODYNAMICS 


Starting  with  the  heat  supplied  by  the  hot  body  we 
have  first  the  unavoidable  losses  even  under  ideal  conditions 
which  according  to  the  second  law  of  thermodynamics 
must  accompany  any  transformation  of  heat  into  mechan- 
ical energy.  Of  the  heat  ideally  transformable  into  work 
only  a  part  is  transmitted  to  the  piston  owing  to  unavoidable 
cylinder  losses.  Of  the  work  actually  reaching  the  piston 
only  a  part  is  delivered  by  the  engine  due  to  the  mechanical 
friction  in  the  mechanism. 

These  conditions  may  be  represented  as  follows: 


Delivered 

Ideally 
available 

Indicated 
work 

work 
Mechanical 

eat 

work 

Cylinder 

losses 

supplied 

losses 

Heat 

necessarily 
degraded 

1  Cycle 

losses 

The  ideally  available  work  may  be  based  upon  the  work 
obtainable  from  a  Carnot  cycle  operating  under  the  con- 
ditions of  the  given  engine.  It  is  however  more  usual 
to  use  the  Rankine  cycle  with  complete  expansion  as  the 
standard  cycle  for  vapor  engines. 

If    q  represents  the  heat  absorbed  per  pound  of  vapor 
used  in  the  engine, 
qr,  the    heat    transformed    into    work    per    pound    of 
vapor  by  an  ideal  Rankine  cycle  under  the  con- 
ditions under  which  the  engine  runs, 
Qa,  the  actual  heat  transformed  into  work  per  pound 


VAPOR   CYCLES  211 

of  Steam  supplied  to  the  cylinder  of  the  actual 
engine  as  determined  by  an  indicator, 
Zi'b,  the   work   obtained   at    the   brake   per    pound   of 
steam  supplied  to  the  engine, 
we  have  as  defining  equations  the  following, 

ideal  thermal  efficiency  =cvcle  efficiency=77r  =  — , 

1 

actual  thermal  efficiency  =  indicated  thermal  efficiency = 77a  =  —  , 

9 

cylinder  efficiency  —y]c  =  —, 

mechanical  efficiency      =  7?m  =  -y- . 

Jqa 

Exercise  211.  (a)  Show  that  the  actual  thermal  efficiency 
equals  rjc-vr- 

(b)  Show  that  the  total  engine  efficiency  equals  vt-vc-  vm- 

If  nir  represents  the  pounds  of  vapor  required  per  indi- 
cated horse-power  hour  with  the  Rankine  cycle,  and 

iria  represents  the  pounds  of  \apor  required  per  indicated 
horse-power  hour  under  the  conditions  existing  in  the  actual 
engine, 

the  heat  transformed  into  work  per  indicated  horse-power 
hour  =  qriflr  =  qania- 

As  the  cylinder  efficiency  is  defined  as 


it  follows  that 

Mr 


212  THERMODYNAMICS 

Exercise  212.  An  engine  supplied  with  steam  at  115  pounds 
per  square  inch  absolute  and  containing  one  per  cent  of  moisture 
was  found  to  use  21  pounds  of  steam  per  indicated  horse-power 
hour  when  the  temperature  in  the  condenser  was  140°  F.  Com- 
pute 

(a)  the  cylinder  efficiency, 

(b)  the  heat  supplied  per  i.h.p.  per  minute, 

(c)  the  actual  thermal  efficiency  of  the  engine. 

Exercise  213.  An  engine  test  showed  the  i.h.p.  to  be  500 
and  the  heat  supplied  per  i.h.p.  per  minute  to  be  290  B.t.u. 
with  a  boiler  pressure  of  170  pounds  per  square  inch  gage,  quahty 
of  steam  0.98,  vacuum  27  inches,  and  the  barometer  at  31 
inches.     Compute 

(a)  the  ideal  thermal  efi5ciency, 

(b)  the  ideal  and 

(c)  the  actual  steam  consumption, 

(d)  the  cylinder  efficiency, 

(e)  the  actual  thermal  efficiency. 

Exercise  214.  An  engine  has  a  cylinder  efficiency  of  0.60, 
a  mechanical  efficiency  of  0.85,  and  operates  under  conditions 
for  which  the  Rankine  cycle  would  transform  200  B.t.u.  per 
pound  of  steam  into  work.  What  is  the  steam  consumption 
of  this  engine  per  brake  horse-power  hour? 


CHAPTER  XIII 

FLOW  OF   FLUIDS 

Section  XXXVIII 

FUNDAMENTAL    EQUATIONS 

In  the  following  discussions  the  particles  of  the  flowing 
fluid  are  assumed  to  move  along  stream  lines.  Any  surface 
composed  of  stream  lines  and  enclosing  a  portion  of  the 
flowing  fluid  may  be  regarded  as  a  tube  thru  which  the 
fluid  flows.  Note  that  no  fluid  passes  thru  the  walls  of 
this  imaginary  tube. 

Steady  flow  is  also  assumed.  This  means  that  at  any 
fixed  point  on  any  stream  line  the  pressure,  the  specific 
volume,  the  temperature,  the  velocity,  and  in  the  case  of 
wet  vapors,  the  quality  of  the  fluid  remain  constant. 

As  shown  in  Fig.  6i  consider  a  portion  of  the  moving 
fluid  bounded  by  stream  lines  and  by  two  sections  (marked 
I  and  2)  normal  to  these  stream  lines.  These  sections  may 
be  assumed  to  be  plane. 

Let  the  following  notation  apply  to  any  section,  such 
as  I  or  2. 

W  is  the  weight  of  fluid  passing  any  section  per  second 

in  pounds  per  second, 
V,  the  volume  per  unit  weight  of  fluid  at   this  section, 
in  cubic  feet  per  pound, 
213 


214  THERMODYNAMICS 

w,  the  velocity  of  the  fluid,  in  feet  per  second  at  this  same 

section  for  which 
A,  represents  the  area  in  square  feet. 
Then  the  volume  of  fluid  passing  this  section  per  second 
equals  both  Aw  and  Wv  so  that 

Aw=Wv. 

As  no  fluid  accumulates   between   sections    the   weight 
of  the  fluid  passing  any  section  must  be  constant,  so  that 

W  =  —  =  — — ^-  =  a  constant. 

V  Vi 

This  is  the  equation  of  continuity,  the  first  of  the  two 
fundamental  equations  governing  the  flow  of  fluids. 
1. 


Fig.  6i. 

The  second  fundamental  equation,  a  special  form  of 
the  law  of  conservation  of  energy,  will  now  be  deduced. 

The  various  forms  of  energy  which  may  be  present 
during  a  flow  from  section  i  to  section  2  (Fig.  61)  are 
kinetic  energy,  energy  due  to  pressure,  potential  energy 
due  to  gravitation,  internal  energy  of  the  fluid,  and  heat 
energy  which  may  be  supplied  to  the  flowing  fluid  during 
its  passing  from  section  i  to  section  2. 


FLOW   OF   FLUIDS 


215 


Let  p  represent  the  absolute  pressure  at  any  section, 
h,  the  position   of  that   section   with   reference   to  any 

horizontal  datum  plane, 
n,  the   internal  energy  of  one   pound   of  fluid   at   that 
section,  in  heat  units, 
then  the  kinetic  energy  of  one  pound  of  the  fluid  is 

•up' 

The  energy  of  one  pound  of  the  fluid  due  to  the  existing 
pressure  is 

P, 

because  as  one  pound  of  fluid  occupies  v  cubic  feet  it  could 
displace  a  piston  (area  A  square  feet)  thru  —  feet  at  the 
same  time  exerting  a  force  of  Ap  pounds  and  therefore 
perform  {Ap){  —  )  =pv  foot-pounds  of  work. 

The  potential  energy  due  to  gravitation  of  one  pound  of 
the  fluid  with  reference  to  the  datum  plane  would  be  h 
foot-pounds. 

The  changes  in  these  various  forms  of  energy  in  one 
pound  of  fluid  due  to  its  transfer  from  section  i  to  section 
2  (Fig.  61)  are 

,       p2'V2  —  p\V\,       112  — ll\ 

H 

and  at  the  same  time  the  change  in  the  internal  energy 
of  the  fluid  is 

«2— «i- 


2l6  THERMODYNAMICS 

The  sum  of  the  changes  in  all  these  forms  of  energy 
must  equal  the  heat  which  has  been  supplied  from  some 
outside  source  to  the  flowing  fluid  during  its  motion  from 
I  to  2,  let  this  heat  per  pound  of  fluid  be  denoted  by  152, 
then 

\-{p2V2  —  piVi)-\-{h2  —  hi)+J{u2  —  ui)^Jiq2 

which  is  the  energy  equation. 

Introducing  the  concept  of  the  heat  content  i,  defined 
by  the  equation 

the  energy  equation  becomes 

'^''~'^^%(fe-/^i)+/fe-n)  =  /ig2. 

It  will  be  of  interest  to  note  that  if  no  heat  is  supplied 
(192  =  0)  and  if  no  change  in  the  internal  energy  occurs 
(w2  =  wi)  the  energy  equation  reduces  to 

■ \-  {p2V2  —  p\V\)-\-  {112  —  h\)  =  O. 

This   is   Bernoulli's   equation   as   used  in  hydraulics.     For 

water  flowing  under  these  conditions  z'2  =  i'i  =  ^ — ,  and  we 
°  62.4 

have 

•W2^  ,      p2      .    J  Wi^         pi  J 

2g        62.4  2g        02.4 

In  the  derivation  of  the  above  energy  equation  no 
mention    is   made   of   the   effect   of   frictional   resistances. 


FLOW   OF   FLUIDS  217 

These  would  decrease  the  final  kinetic  energy  owing  to 
the  transformation  of  mechanical  energy  into  heat.  This 
heat  would  either  remain  in  the  fluid  or  pass  to  some  other 
body  of  the  system.  Thus  the  frictional  resistances  would 
not  change  the  sum  of  the  energies,  but  would  effect  a  redis- 
tribution of  energy  among  the  various  kinds  present  in  the 
system. 

Another  form  of  the  energy  equation  in  which  the 
effect  of  the  frictional  resistance  appears  is  derived  as 
follows. 

In  the  fundamental  equation 


dq  =  du-\-—pdv, 


in  which  we  may  replace  w  by  i  by  means  of  the  relation 


and  obtain  da  =  di ^, 


q  represents  all  the  heat  received  by  the  substance  either 
from  an  external  source  or  generated  by  frictional  resistance 
from  mechanical  energy  within  the  substance  itself. 

Let  1^2  be  the  heat  energy  received  from  external  sources 
during  the  flow  from  i  to  2,  Fig.  61,  and  1/2  the  mechanical 
energy  which  disappears  during  this  flow  to  reappear  as 
heat,  then  we  have 


I  I  C^'' 

192  +  -^  1/2  =  {h  -h)-  -J  J    vdp. 


2l8  THERMODYNAMICS 

But  from  the  energy  equation  already  deduced 

J{i2-ii)= {h2  —  hi)+Jiq2, 

therefore 

I  I  W')   — Wi         I  I    C^^ 

vdp. 


.2_,„,2 


W2   —  Wi 

or  

2g 


Section  XXXIX 
ADIABATIC,   FRICTIONLESS   FLOW 

In  most  practical  examples,  at  least  as  a  first  approxi- 
mation, the  flow  may  be  considered  frictionless  and  no 
heat  may  be  assumed  to  be  exchanged  between  the  fluid 
and  any  external  body. 

Under  these  conditions  1/2  =  0  and  1(72  =  0  and  the  change 
of  state  becomes  isentropic. 

The  equation  of  continuity  is  still 

W  =  —  =  a  constant, 

V 

and  the  energy  equations  reduce  to 

^\^= ^ =  J(ll-l2), 

\2g/  2g 


and  ^w:\     "^-w, 


2  /'Ps 

-=-       vdp. 

Jpi 


2g/  2g 

The  term  {ho  — In)  is  omitted  as  the  difference  in  heads  is 
usually  small  in  practical  applications  to  vapors  and  gases. 


FLOW   OF   FLUIDS  219 

The  second  form  of  the  energy  equation  is  so  important 
that  it  may  be  well  to  explain  its  derivation  from  the  first 
form  under  the  special  conditions  of  reversible  adiabatic 
change  of  state. 

Under  these  conditions,  by  reason  of 

as  Aq  =  o 

we  have  An  =  —  Aw 

pdv. 
"1 

Thus  the  energy  equation 

Ag)=/(.-.-.y 

or  as  it  may  be  written 

A{—j=J{ui-H2)-\-plVl  —  p2V2 

becomes  A I  —  1=  I    pdv-\- piv\  — p2V2. 

In  Fig.  62,  point  i  represents  the  state  of  the  fluid  existing 
at  section  i,  Fig.  61,  and  point  2  the  state  existing  after 
flow  to  section  2,  Fig.  61.  Under  the  special  conditions 
the  process  i  2  is  reversibly  adiabatic  and  the  area  under 

pdv.     Also   the    shaded   area,   Fie;. 
62,  represents    I    pdv-\- piv\  — P2V2    which    in    turn    equals 
^^  up  or -^^^vdp, 

vdp. 

n. 


thus  A 


2g 


220 


THERMODYNAMICS 


Remember  that  during  reversible  adiabatic  flow  the 
change  in  kinetic  energy  of  the  flowing  fluid  equals  the  work 
represented  by  the  area  behind  the  curve  representing  the 
process  on  the  pv-plsine,  the  shaded  area  in  Fig.  62. 


Exercise  215.  Show  that 


ipiVi-p2V'^, 


where  m=k=  1.40  for  ideal  (diatomic)  gases, 

w=  1.035  +0.100x1,  for  wet  steam  with  an  initial  quality  Xi, 
and     m— 1.2,1  for  superheated  steam. 


Fig.  62. 


Exercise    216.  Starting   with    d{  —  I  =/(/i  — ja),    show   that 
for  adiabatic  flow 

for  ideal  gases, 


a(  — )=  7(9/-. r,rO-/(g2'+X2r2) +(/»!- />2)f  32 


for  wet  vapors, 


2g 


■■Jqi  —  J{q2—XiH)  +(pi  —  p2)vs2 


for  superheated  vapors  changing  to  wet  vapors  during  the  flow. 


FLOW   OF  FLUIDS 


221 


Exercise  217.  If  the  initial  veiocity  of  one  pound  of  steam 
is  zero  and  the  conditions  under  which  expansion  occurs  are 
such  as  to  render  300  B.t.u.  available  for  the  production 
of  velocity,  what  will  be  the  velocity  of  the  steam  _after 
expansion? 

Solve  this  problem  by  means  of  the  equations  deduced  above 
and  check  the  result  by  means  of  the  velocity  scale  on  the  Mollier 
diagram. 

Section  XL 
FLOV/  THRU   AN   ORIFICE 


A  very  short  channel  whose  section  continually  decreases 
in  the  direction  of  the  flow  so  that  the  last  section  has  the 
smallest  area  is  called  an  ori- 
fice. The  area  of  the  orifice 
is  the  least  sectional  area  of 
the  channel.  The  flow  thru 
an  orifice  may  be  regarded  as 
both  adiabatic  and  frictionless, 
owing  to  the  small  suifaces  of 
contact  and  to  the  short  time 
the  fluid  requires  to  pass  thru 
the  orifice. 

Let   the    state   of   the    fluid 
on  the   up-stream   side   of  the 

orifice  be  represented  by  p\  and  v\  at  a  point  where  its 
velocity  is  practically  zero  (Fig.  63),  also  let  Ao  be  the  area 
of  the  orifice,  Wq,  the  velocity  of  flow  at  the  section  whose 
area  is  Ao,  and  p2  and  V2,  the  state  of  the  fluid  on  the  down- 
stream side  of  the  orifice. 


Fig.  63. 


222  THERMODYNAMICS 

From  the  energy  equation 

we  find  '^''^^^^^i^''''Y~{yJ  "  I' 

where  m  is  the  exponent  of  the  equation  of  the  adiabatic 
process  pv"*  =  piVi"^. 

Exercise  218.  Deduce  the  value  of  Wq-  given  above. 

From  the  equation  of  continuity 

T„    Aw    AoWo 

if  we  assume  that  the  pressure  and  the  specific  volume  of 
the  fluid  at  the  mouth  of  the  orifice  equal  the  pressure  and 
uhe  specific  volume  on  the  down-stream  side  of  the  orifice 
we  find  that  the  pounds  of  fluid  passing  the  orifice  per  second 
equal 


^    ""m-i  vil\pi/  \pi 


Exercise  219.  Deduce  this  value  of  W. 

As  a  special  case  of  the  abo\'e  equations  assume  the 
pressure  on  the  down-stream  side  of  the  orifice  to  be  main- 
tained at  absolute  zero,  so  that  p2  =  o.  If  we  also  assume, 
as  the  conditions  seem  to  justify,  that  the  pressure  at  the 
section  whose  area  is  ^o  is  also  p2  =  o  we  find  the  velocity 

at  this  section  to  be 

m 

Wo=2g pivi 

m—i 

and  the  discharge  in  pounds  per  second  to  be 

W=o. 


FLOW   OF   FLUIDS  223 

This  last  conclusion  can  surely  not   agree  with  the  facts 
especially  as  the  velocity  attained  by  the  fluid  is  not  zero. 

Further  investigation  of  the  expressions  for  Wo  and  W 
shows  that  as  the  pressure  on  the  down-stream  side  of  the 
orifice  is  made  gradually  less  than  pi  the  value  of  Wo  steadily 
increases,  which  is  to  be  expected,  but  the  value  of  W 
at  first  increases,  reaches  a  maximum  when  the  pressure 
on  the  down-stream  side  has  been  diminished  to 


,=Pi{, 


w+i 


and  then  decreases  with  diminishing  down-stream  pressure 
until  it  becomes  zero  when  p2  becomes  zero.  The  par- 
ticular pressure  pc  which  yields  a  maximum  discharge  is 
called  the  critical  pressure. 

Exercise  220.  Deduce  the  value  of  the  critical  pressure 
given  above. 

In  1839,  Saint  Venant  and  Wantzel  advanced  the  hy- 
pothesis that  the  pressure  at  the  mouth  of  an  orifice  is  not 
always  equal  to  the  pressure  on  the  down-stream  side  of 
the  orifice,  that  is  that  po  is  not  always  equal  to  p2  and 
that  for  values  of  p2<pc,  po  no  longer  equals  p2  but  re- 
mains constant  and  equal  to  pc. 

Exercise  221.  The  velocity  of  sound  in  any  fluid  equals 

'\'  gmpv, 

where  g  is  the  acceleration  due  to  gravity, 

m  is  the  exponent  of  adiabatic  expansion, 

p  is  the  pressure, 

V  is  the  specific  volume, 


224 


THERMODYNAMICS 


the  last  three  of  the  fluid  under  the  conditions  existing  when 
the  sound  is  transmitted.  Show  that  the  greatest  possible 
velocity  attainable  by  a  fluid  flowing  thru  an  orifice  equals 
the  velocity  of  sound  in  this  fluid  when  in  the  state  existing 
at  the  mouth  of  the  orifice. 

The  conclusion  reached  in  Exercise  221  indicates  a  pos- 
sible explanation  for  the  existence  of  the  critical  pressure. 
Assume  p2<pc  then  the  jet  issuing  from  the  orifice  at  a 
pressure  greater  than  p2  and  being  surrounded  by  fluid 
under  lesser  pressure  explodes  and  the  pressure  in  the 
jet  is  suddenly  reduced  to  p2-  This  pressure  p2  is  prop- 
agated toward  the  mouth  of  the  orifice  with  the  velocity 
of  sound  but  the  transmission  of  this  pressure  p2  up  to 
the  mouth  of  the  orifice  is  prevented  by  the  outrush  of 
the  fluid  which  also  moves  with  the  velocity  of  sound. 

Thus  according  to  the  hypothesis  of  Saint  Venant  and 
Wantzel,  which  has  been  fully  confirmed  by  experiment, 
two  cases  arise  during  the  flow  of  fluids  thru  an  orifice, 
namely 

(i)  when  p2>pc, 

(2)  when  p2^pc- 

In  the  first  case  {p2  >  pc) 


ni—i         I        \pi 


m—  \ 
pp\     m 


and  IF=^o(f^)"V^«^^  ^i 


-{f)    \ 


the  values  depending  upon  the  back  pressure. 


FLOW   OF   FLUIDS  225 

In  the  second  case  {po  <  pc) 


'^o=\\2g 7—p\Vi 


and  w  =  aJ^Y"'J^J^, 

\m-\-i/        \m-]-i\  vi 

which  values  are  independent  of  the  back  pressure. 

Exercise  222.  Deduce  the  values  of  Wo  and  W  given  above. 

Exercise  223.  Show  that  the  critical  ratio  equals  0.528  for 
air  and  0.577  for  dry  saturated  steam. 

Exercise  224.  Dry  saturated  steam  escapes  to  the  atmosphere 
thru  an  orifice  (area  0.3  square  inch)  from  a  boiler  in  which 
the  pressure  is  maintained  at  150  pounds  per  square  inch  abso- 
lute.    Find  the  velocity  of  discharge 

(a)  by  means  of  the  above  theory, 

{b)  by  means  oi  w=  v  2g/(/i  — z^), 

(f)  by  means  of  the  MoUier  diagram. 

Exercise  225.  What  would  be  the  velocity  of  discharge  in 
Exercise  224  if  the  whole  fall  in  pressure  could  be  utilized  in 
producing  velocity? 

The  Discharge  thru  an  Orifice. — The  discharge  thru 
an  orifice  may  be  computed  in  any  case  for  any  fluid  by 
means  of  the  equations  given  above,  provided  the  value 
of  m  is  known.  Two  well-known  formulas  which  are 
special  cases  of  the  more  general  formulas  already  given 
will  now  be  deduced. 

Fliegner's  formula  for  air,  for  the  discharge  thru 
an  orifice  when  p2<pc,  in  terms  of  the  initial  pressure 
Pi  and  the  initial  temperature  Ti  of  the  air  may  be 
derived  from 


W  =  Ao 


2      \m-l      /  2gm       I  pi 

m-{-i)        \m-\-i\v\ 


2  26  THERMODYNAMICS 

by  eliminating  Vi  by  means  of  the  relation 
piVi  =  RTi. 
In  this  way  we  obtain 


\m-\-i/        \m-{-i\RTi 


and  as  m  =  k=i.4o,    and    R—S3-3 

where  W  is  the  discharge  of  air  in  pounds  per  second, 
Ao,  the  area  of  the  orifice  in  square  feet, 
Pi,  the  pressure  in  pounds  per  square  foot, 
Ti,  the  absolute  temperature  in  degrees  F, 
the  last  two  on  the  up-stream  side  of  the  orifice. 

As  the  critical  ratio  for  air  is  0.528,  approximately  c.5, 
this  formula  for  W  can  only  be  used  for  the  discharge  into 
the  atmosphere  when  the  reservoir  pressure  is  at  least 
twice  the  atmospheric  pressure.  Under  these  conditions 
the  formula  agrees  with  the  results  of  Fliegner's  experiments. 
For  the  discharge  of  air  from  reservoirs  in  which  the 
pressure  is  less  than  twice  the  atmospheric  pressure  into 
the  atmosphere  Fliegner  proposed  the  empirical  formula 


^^'^ 


W=i.o6A.    /.''(/'i-^) 


Ti 


where  pa  is  the  atmospheric  pressure  in  pounds  per  square 
foot.  This  formula  may  be  used  instead  of  the  more 
cumbersome  formula  given  on  page  224. 


FLOW    OF   FLUIDS 


227 


Grashof  s  formula   for  wet  steam  when  the  discharge 

occurs  under  the  condition  p2<pc, 

,  J      2      \m-l 

where  Pc=pi] — ; — 

[■m-f-i 

and  m=  1.035+0.100.V1 

Pc 

so  that  for    a*i  =  i.o,    ^  =  1.135,    and    —  =  0.577 

p\ 

rt;i  =  o.9,    m=i.i25,     and     —  =  0.580 

pi 

0:1  =  0.8,    w=i.ii5,     and     —  =  0.582 

pi 

.Yi  =  o.7,    w=i.io3,     and     ^  =  0.584 

is  based  upon  the  relation 

Pi(vi"y  =  C  =  SL  constant 

between  the  pressure  and  the  specific  volume  of  dry  sat- 
urated steam  for  which  5  may  be  assumed  to  be  1.063 
and  for  which 

C=  14.7X  144X  (26.79)1-063  =  69,600. 

By  means  of    this  relation  Vi  may  be  eliminated  from 
the  formula  for  W.    As  .ri  is  the  initial  quality  of  the  steam, 

fl  =  -Vl"c'l"+(l-.Vl)l'l' 

and  as  the  quality  xi  is  usually  high  and  the  pressure  never 
excessively  great  we  may  assume  as  an  approximation 


2  28  THERMODYNAMICS 

Thus  p,(vi"y=pi(z-J=C 

or  vi  =  xii — 1    ,  approximately. 

\Pi/ 

Also  from  m=  1.035+0.100x1,  as  xi   remains  near  one  we 
have  approximately 

m=  1.135. 

Substituting  these  values  of  vi  and  m  in  the  formula 


\m-\- 1/        \  m-\- 1  \  J^i 


we  obtain        W  =  o.oig    °  , —   , 

V  xi 

where  y4o  is  measured  in   square  feet,  pi   in  pounds  per 
square  foot,  and  W  in  pounds  per  second. 

If  Ao  is  ex-pressed  in  square  inches,  pi  in  pounds  per 
square  inch  and  W  in  pounds  per  second  then 

TF  =  o.oi65^^^^. 

V.V1 

Exercise  226.  Deduce  Grashof's  formula. 
Napier's  formula  for  dry  saturated  steam  discharging 
under  the  conditions  ^2<o.58/'i  is 

70 

where  ^0  is  to  be  measured  in  square  inches,  pi  in  pounds 
per   square  inch   and   W  in   pounds  per   second.     It   was 


FLOW   OF   FLUIDS  229 

deduced  by  Rankine  from  experiments  by  Napier.  This 
formula  altho  not  as  accurate  as  Grashof's  is  still  used  on 
account  of  its  simplicity. 

The  discharge  of  a  vapor  thru  an  orifice  can  most 
accurately  be  computed  directly  from  the  fundamental 
equations  

Wo  =  V2gJ{ii  —  io), 

and  W  =  ^^. 

Vo 

When  the  back  pressure  is  less  than  the  critical  pressure 
(p2<pc)  it  must  be  remembered  that  Pc  and  not  p2  must 
be  used  in  finding  to  and  I'o-  Both  Zo  and  Vq  can  most 
readily  be  found  from  the  vapor  diagrams. 

Exercise  227.  Compute  the  discharge  in  pounds  per  minute 
under  the  conditions  described  in  Exercise  224,  by  means  of 

(a)  Napier's  formula, 

(b)  Grashof's  formula, 
(f)   the  steam  diagrams. 

The  formulas  for  the  flow  of  vapors  deduced  in  this 
section  are  strictly  applicable  only  to  the  flow  of  vapors 
initially  slightly  wet.  The  moisture  must  be  present  in 
very  small  drops  distributed  uniformly  thruout  the  mass 
of  vapor  so  as  to  approximate  a  homogeneous  condition. 
These  drops  of  liquid  seem  to  serve  as  nuclei  or  centers  of 
condensation  and  become  larger  as  the  quality  of  the 
flowing  vapor  decreases.  If  the  vapor  supplied  is  dry 
saturated  or  superheated  it  is  found  that  the  vapor  does 
not  change  its  quality  with  the  expansion  according  to 
the  theoretical  indications  but  remains  in  a  supersaturated 
condition  owing  to  the  lack  of  centers  of  condensation 
which  seem  necessary  to  start  condensation. 


230  THERMODYNAMICS 

If  the  liquid  has  separated  from  the  vapor,  if  it  is  not 
uniformly  distributed  thruout  the  vapor  as  fog,  it  does 
not  serve  the  purpose  above  described.  It  then  simply 
flows  along  the  walls  of  the  orifice  with  much  smaller  velocity 
than  the  issuing  vapor  and  the  formulas  deduced  do  not 
apply.  The  liquid  under  these  conditions  partially  ob- 
structs the  orifice,  lessens  the  discharge  and  if  present  in 
considerable  amount,  causes  a  sneezing  action  during  dis- 
charge. 

For  further  information  on  this  subject  the  reader  is 
referred  to 

Callendar:  On  the  Steady  Flow  of  Steam  thru  a 
Nozzle  or  Throttle,  Journal  I.  Mech.  E.,  p.  53,  February, 
1915;  and 

Leblanc:  Machine  Frigorifique  a  Vapeur  D'Eau  et  a 
Ejecteur.     Gauthier-Villars,  Paris,  191 1. 

Section  XLI 

FLOW    THRU    A    NOZZLE    INCLUDING    THE    EFFECT   OF 
FRICTION 

As  shown  in  the  last  section  the  velocity  attained  by  a 
fluid  flowing  thru  an  orifice  increases  with  decreasing 
back  pressure  until  the  back  pressure  attains  a  value 
equal  to  the  critical  pressure 


Pi 


/      2      \m-l 

\W+I/ 


and  then  remains  constant  irrespective  of  any  additional 
decrease  in  back  pressure. 

Under  these  conditions  the  energy    I    vdp  which  is  lib- 


FLOW   OF   FLUIDS  23 1 

erated  by  the  drop  in  pressure  cannot  be  wholly  converted 
into  kinetic  energy  of  the  flowing  fluid  when  the  back 
pressure  is  less  than  the  critical  pressure. 

A  nozzle  is  a  tube  of  such  form  that  the  fluid  flowing 
thru  it  attains  the  full  velocity  and  therefore  kinetic 
energy  which  may  be  expected  from  a  given  drop  in  pressure. 

The  Form  of  a  Nozzle. — Let  us  investigate  the 
variation  in  the  cross-section  of  a  stream  under  frictionless, 
adiabatic  flow  during  which  the  pressure  gradually  falls 
from  an  initial  pressure  pi  to  a  final  pressure  p2  and  during 
which  the  velocity  increases  from  zero  to  W2  in  accordance 
with  the  law 

w^—wr     r^'  , 
=  I     vdp 

no  matter  how  low  p2  may  be  made. 

Under  these  conditions  w  continuously  equals 


m  —  1 

and  the  varying  cross-section  of  the  stream 

A=  — , 

w 


where  W  is  a  constant. 
As  pv'"  =  pivi'" 


.P 


1 

Pi" 


1 


Mr       I  pi 


so  that  A  = 


^^,p4-{i-y] 


232  .  THERMODYNAMICS 

To  investigate  the  variation  of  A  as  the  pressure  p  falls 
we  first  seek  maximum  and  minimum  values,  by  placing 

dA 
dp 


As  A  = 


JL  _J. 

Wvipi""  p   "» 


^  m—i 


any  value  of  p  that  makes  A   a  maximum  or  a  minimum 
will  also  make 

2  1  — J7»    m  +  1    _1_ 

=  {P^-Pl'^p   "^    \      2 

2  1  — W»     CT+1 

or  even  ^"  =  />^-/^i~^/?^^ 

a  maximum  or  a  minimum. 
Thus  the  equation 

dp      m'^  \   m    / 

yields  as  a  critical  value 

p  =  pi 


w+i 


The  cross-section  of  the  stream  will  thus  reach  a  maxi- 
mum or  a  minimum  value  when  the  pressure  in  the  fluid 
drops  to  the  critical  value  just  found.  Let  the  critical 
pressure  be  denoted  by  pc,  the  corresponding  velocity  and 


FLOW    OF   FLUIDS  233 

specific  volume  of  the  fluid  by  Wc  and  Vc  respectively,  and 

the    corresponding    cross-sectional    area    by    Ac,   then  the 
velocity  of  the  fluid  at  the  critical  section  is 


Wc 


=  \h-  7~pin  =  V  gmpcVc, 


the  velocity  of  sound   in   the  fluid  under   the  conditions 
existing  at  this  section,  see  Exercise  221. 

In  order  to  demonstrate  that  Ac  is  a  minimum  cross- 
section  of  the  stream  it  will  be  more  convenient  as  well 
as  more  profitable  to  develop  and  study  another  form  of 

-r—  instead  of  using  —^  which  may  be  obtained  from  the 
dp  "  dp-  ^ 

dA 
value  of  -z—  already  found. 

The  fundamental  equations  governing  the  flow  under 
consideration  are 

^(jgj^-'^P ^'^ 

Aw=Wv (2) 

and  />!>"'=  a  constant (3) 

As  we  desire  to  find  the  change  in  area  of  the  cross- 
section  due  to  a  change  in  pressure  we  must  express  dA 
in  terms  of  dp.     To  obtain  dA  differentiate  equation  (2) 

■wdA-{-Adw=Wdv (4) 

Next  eliminate  dv  and  dw  and  introduce  dp.    From  equation 

■(I)  dw=-^-^ 

w 

and  from  equation  (3)  or  its  equivalent 


234  THERMODYNAMICS 

log  p-\-7n  log  v  =  \og  (a  constant) 

dp  ,      dv 
—^-\-m —  =  o 
p  V 

or  dv= -dp. 

in  p 

Substituting  these  values  in  equation  (4)  we  find 

wdA ^ — -= dp 

w  ni  p 

dA     Aev      Wv 
dp      w^      wmp 

Eliminating  W  by  means  of  equation  (2)  we  obtain 

dA  _A{ gmpv—w^ } 
dp  mpw^ 

Note  that  gmpv  is  the  square  of  the  velocity  of  sound 
at  any  section  A  at  which  the  velocity  of  the  fluid  is  w 
in  a  medium  whose  pressure  and  specific  volume  are  p 
and  V. 

All  quantities  in  the  right-hand  member  of  this  equation 
are  necessarily  positive;  negative  values  of  these  quantities 
are  excluded  by  our  physical  conception  of  the  problem. 
However  dp  is  always  negative,  for  the  pressure  is  assumed 
to  decrease  continually  from  the  section  Ai  to  the  section  A2' 

Let  us  now  study  the  signs  of  dA.  This  sign  will  deter- 
mine whether  the  cross-section  of  the  stream  is  increasing 
or  decreasing  at  any  point. 

(i)  As  dp  is  never  zero,  dA  is  zero  when 

w=\  gmpv. 


FLOW   OF   FLUIDS  235 

That  is,  the  cross-section  of  the  stream  is  neither  increasing 
nor  decreasing,  when  the  velocity  of  the  fluid  equals  the 
velocity  of  sound  in  the  fluid  under  the  conditions  existing 
at  the  section  considered. 

(2)  As  dp  is  negative,  dA  is  negative  when 

w  <  \  gnipv. 

That  is,  the  cross-section  must  decrease  during  that  portion 
of  the  flow  for  which  the  velocity  of  the  fluid  is  less  than 
the  velocity  of  sound  in  the  fluid  at  the  section  considered. 

(3)  As  dp  is  negative,  dA  is  positive,  when 

w^w  gmpv. 

That  is,  the  cross-section  of  the  stream  must  increase 
during  that  portion  of  the  flow  for  which  the  velocity  of 
the  fluid  is  greater  than  the  velocity  of  sound  in  the  fluid 
at  the  section  considered. 

Therefore  a  nozzle  in  which  a  continuous  decrease  in 
pressure  with  a  continuous  increase  in  velocity  is  to  occur 
irrespective  of  any  limiting  pressures  must  at  first  diminish 
in  section  from  a  very  large  to  a  minimum  section  (called 
the  throat  of  the  nozzle)  and  then  the  section  must  con- 
tinually increase  with  increasing  velocity  and  decreasing 
pressure. 

The  length  of  the  whole  nozzle  should  be  made  as  short 
as  possible  so  as  to  keep  down  the  losses  due  to  friction. 
The  portion  of  diminishing  sectional  area  can  be  made 
very  short,  the  section  at  first  diminishing  rapidly  and 
then  more  slowly  up  to  the  throat  of  the  nozzle.  For 
practical  reasons  of  manufacture  an  extremely  short  length 
of  the  nozzle  at  the  throat  is  usually  made  cylindrical. 


236  THERMODYNAMICS 

The  portion  of  increasing  sectional  area  cannot  be  made 
as  short  as  the  portion  of  diminishing  sectional  area  for 
if  the  section  of  the  nozzle  enlarges  too  rapidly  it  is  found 
that  the  stream  no  longer  fills  the  nozzle,  no  longer  follows 
the  wall,  and  therefore  ceases  to  expand  in  the  desired 
manner.  The  diverging  portion  of  the  nozzle  is  commonly 
made  conical  in  form  and  the  angle  of  the  coneshould 
be  about  6°. 

Design  of  a  Nozzle,  Neglecting  Friction. — The  pressure 
at  the  throat  of  a  nozzle  is 


pc  =  p\ 


the  velocity  at  the  throat  may  be  computed  by  means  of 


Wc  =  ^  2gJ{i\  —  ic) 

or  it  may  be  read  directly  from  the  velocity  scale  on  the 
Mollier  diagram.  Note  that  4  is  the  heat  content  corre- 
sponding to  the  critical  pressure  pc  after  a  reversible  adia- 
batic  expansion  from  the  initial  state. 

The  sectional  area  at  the  throat  is  fixed  by  the  number 
of  pounds  of  fluid  that  must  be  discharged  per  second. 
From  the  equation  of  continuity  we  have 

Ac= , 

Wc 

where  Vc  is  the  specific  volume  of  the  fluid  under  the  con- 
ditions existing  at  the  throat. 

The  exit  area  of  the  nozzle  is  found  in  the  same  manner 
by  means  of  

and  Wv2  =  A2'W2, 


FLOW   OF   FLUIDS  237 

where  the  subscripts  2  refer  to  the  conditions  existing  at 
the  end  of  the  nozzle. 

If  the  nozzle  is  to  deliver  energy  at  a  certain  rate  in 
the  form  of  kinetic  energy  the  pounds  of  fluid  passing 
thru  the  nozzle  are  computed  as  follows, 

\Y^  (horse-power)  (550) 
J{h  —  i2) 

Exercise  228.  A  nozzle  is  to  deliver  5  horse-power  in  the 
form  of  kinetic  energy  of  the  moving  fluid.  It  is  to  be  supplied 
with  steam  at  125  pounds  per  square  inch  absolute,  quality 
0.93,  and  is  to  discharge  against  a  pressure  of  0.50  pound  per 
square  inch  absolute.  Compute  the  sectional  area  of  the  required 
nozzle  at  the  throat  and  at  exit,  neglecting  the  effect  of  friction, 

(a)  by  means  of  the  jVIollier  diagram, 

{h)  by  means  of  the  steam  tables. 

The  Ts-plane. — The  solution  of  Exercise  228  by  means  of 
the  steam  tables  involves  the  computation  of  the  quality 
of  the  steam  and  the  use  of  this  quality  in  the  computation 
of  the  drop  in  the  heat  content.  By  means  of  the  graphical 
representation  on  the  T^-plane  formulas  may  be  developed 
in  which  the  drop  in  the  heat  content  is  expressed  in  terms 
of  quantities  given  in  the  vapor  tables. 

A  •  I   /'"''32 

As  i^q+'—j-, 

^\^\=J{H-i-2}^J{q\-q2)  +  {pl-p2)n2. 

If  we  assume  that  the  liquid  line  during  heating  at  the 
constant  pressure  p2  coincides  with  the  corresponding 
portion  of  the  liquid  line  during    heating  at  the   constant 


238 


THERMODYNAMICS 


pressure  pi,  see  page  204,  then  q\  —  q2,  the  shaded  area  Fig. 
64,  may  be  expressed  as  follows  when  x\  is  the  initial 
quality. 

qi-q2  =  qi"-(^yi-x{){Ti-T2)-{q2"-{s2"-S2')T2]. 

Thus 

in  which  each  expression  is  either  given  or  can  be  read 
directly  in  the  vapor  tables. 


f^////////////////^  ' 


(y),U-^P 


K 

dy^ 


Fig.  64. 
Exercise  229.  Show  by  means  of  the  T^-plane  that 


when  the  vapor  is  initially  dry  saturated  and  expands  at  con- 
stant entropy. 

Exercise  230.  Compute  the  energy  available  for  transfor- 
mation into  kinetic  energy  when  one  pound  of  steam  expands 
adiabatically  without  friction  from  200  to  10  pounds  per  square 
inch  absolute,  the  initial  quality  being  0.95 

(c)  by  means  of  the  formidas  just  developed, 

{h)  by  means  of  the  MoUier  diagram. 


FLOW  OF  FLUIDS 


239 


Exercise  231.  Deduce  a  formula  similar  to  the  one  developed 
in  Exercise  229,  but  for  vapors  initially  superheated. 

Exercise  232.  Show  on  the  T^-plane  an  area  representing  the 
energy  available  for  transformation  into  kinetic  energy  during 
an  isentropic  expansion  of  an  ideal  gas  and  from  it  compute 
this  energy.  Compare  the  result  with  the  result  obtained  in 
Exercise  216. 

Adiabatic  Flow  with  Friction. — During  frictionless, 
adiabatic    flow    the   increase  in  kinetic  energy  due  to  a 


Fig.  65. 


given  drop  in  pressure  may  be  represented  on  the  pv-pla.ne 
(Fig.  65)  by  the  area  behind  the  reversible  adiabatic  process 
I  2,  whose  equation  is 


'Iherefore 


|2  — •751,2         /»pi 


W2   —Wi 
2g 


vdp 


(l) 


Under  these  same  conditions  we  have 


2g 


=  J{i\-i2) 


•     (2) 


240 


THERMODYNAMICS 


SO  that  on  the  T^-plane  (Fig.  66)  this  increase  in  kinetic 
energy  is  represented  by  the  heavily  outHned  area  behind 

(pi  ~  p2)VZ2 


the  Hne  i  2,  plus  a  small  quantity 


/ 


.  ■  .  P1J32 

for  by  definition         i  =  q-\-~~ 


and  therefore      ii  —  i2  =  qi  —  q2-\- 


(pl  —  p2)V32 


J 

but  qi  —  q2  is  represented  by  the  area  just  described. 


Fig.  66. 


Now  as  an  extreme  case  conceive  sufficient  frictional 
resistance  during  the  adiabatic  flow  to  destroy  completely 
all  velocity  as  it  develops  then  Wi  =  w  =  W3  and  from 
equation  (2) 

Let  this  final  condition  be  represented  by  the  points  3  in 
both  Figs.  65  and  66  and  the  process  altho  adiabatic 
is  now  one  of  constant  heat  content  represented  by  the 
line  I  3. 


FLOW   OF   FLUIDS  24 1 

Note  carefully  that  the  areas  behind  the  adiabatic,  con- 
stant heat  content  lines  no  longer  represent  either  the  work 
or  the  heat  transformed  into  kinetic  energy.  For  surely 
more  kinetic  energy  would  not  be  obtained  with  the  same 
drop  in  pressure  when  energy  must  be  used  in  overcoming 
the  frictional  resistances. 

The  same  reasoning  applies  to  an  actual  nozzle  for  which 
the  process  lies  between  lines  i  2  and  i  3.  Let  the  process 
be  represented  by  i  4.  This  also  is  an  adiabatic  but  it  is 
now  neither  an  isentropic  nor  a  line  of  constant  heat  con- 
tent.    The  equation 

wi—w^      ^,.       .  . 

still  holds  but  now 

,2 


W4 


.,j2  /»pi 

— [-1/4=  I     vdp, 


where  1/4  is  the  energy  lost  in  friction  between  states  i  and  4 
per  pound  of  fluid,  must  be  used  instead  of  equation  (i) 
given  above,  see  page  217.  Thus  the  area  on  the  ^I'-plane 
behind  the  curve  i  4  not  only  represents  the  mechanical 
energy  which  appears  as  kinetic  energy  but  also  the 
mechanical  energy  re-transformed  by  friction  into  heat. 

If  thru  the  point  4  a  line  of  constant  heat  content  be 
dra\\Ti  until  it  intersects  the  constant  entropy  line  i  2  at 
4',  the  areas  behind  the  line  i  4'  represent  the  kinetic  energy 
available  after  adiabatic  frictional  expansion. 

The  difference  between  the  areas  behind  the  lines  i  2  and 
I  4'  does  not  represent  a  total  loss.  This  energy  is  returned 
to  the  fluid  as  heat  and  causes  the  final  state-point  to  lie 
at  4  instead  of  4'  or  2,  and   therefore  if  a  further   trans- 


242  THERMODYNAMICS 

formation  of  heat  into  mechanical  enregy  is  to  occur  the 
fluid  contains  more  available  heat  energy  than  it  would 
have  contained  if  its  state-point  were  at  2. 
In  Fig.  66  we  have  14  =  14'  so  that 

i2  —  i4=i2  —  i4, 

therefore  the  area  behind  2  4'  and  up  to  the  liquid  line 
is  practically  equal  to  the  area  below  the  line  2  4  and  down 
to  the  axis  of  absolute  zero  temperature. 

Exercise  233.  Deduce  formulas  for  computing  the  quality 
of  a  wet  vapor  leaving  a  nozzle  after  an  adiabatic  expansion 
during  which  y  per  cent  of  the  ideal  heat  drop  is  lost  in  friction. 

Exercise  234.  Sketch  diagrams  similar  to  Figs.  65  and  66 
for  a  vapor  expanding  from  a  superheated  to  a  wet  condition. 

On  the  is-plane,  the  MoUier  diagram,  the"  process 
described  above  on  the  pv-  and  the  T^-planes  would  appear 
as  shown  in  Fig.  67.  The  numbering  corresponds  thruout 
with  that  on  the  preceding  figures. 

As  it  is  impossible  to  locate  the  state-point  4  directly  and 
as  the  process  i  4  is  unknown  it  is  customary  to  compute  the 
difference  between  the  heat  contents  at  i  and  2  assuming 
reversible  adiabatic  change  of  state  (at  constant  entropy 
and  therefore  without  frictional  losses)  and  then  subtract 
a  certain  fraction,  y,  of  this  heat  drop,  i\  —  i2,  so  as  to 
allow  for  the  loss  due  to  friction.  On  the  w-plane  the 
distance  i  2  represents  ii  —  i2  and  the  distance  4'  2  repre- 
sents y{ii—i2)  so  that  the  distance  i  4'  represents 

(i  — y)(ji  — ^2). 
The  final  velocity  of  the  vapor  is  computed  from 

'"^'=J(i-y)(u-W. 


FLOW   OF   FLUIDS 


243 


The  final  quality  of  the  vapor  however  must  be  read 
from  the  point  4  and  not  from  either  4'  or  2. 

Design  of  Nozzles  Including  Friction. —  Computa- 
tions involving  the  flow  of  vapors  thru  nozzles  can 
most  readily  be  performed  by  means  of  the  Mollier  dia- 
gram, Fig.  67.  The  flow  is  first  assumed  reversibly  adia- 
batic  and  the  state-points  i  and  2  are  located,  then  the 
point  4'  is  located  so  that  the  distance  2  4'  is  the  yth.  part 
of  the  distance   i  2  where  y  is  the  given  fraction   of   the 


Fig.  67. 


ideal  heat  drop  that  is  lost  in  friction  as  determined  by 
experiment.  This  fraction  may  vary  from  0.08  to  0.20. 
The  actual  velocity  can  then  be  found  from  the  velocity 
scale  or  by  means  of  the  fundamental  formula. 

From  the  available  heat  drop  and  the  given  conditions 
the  mass  of  fluid  which  must  pass  thru  the  nozzle  can  be 
determined. 

To  find  the  exit  area  of  the  nozzle  the  specific  volume 
of  the  vapor  at  exit  must  be  found  and  this  in  turn  re- 
quires that  the  quality  of  the  vapor  leaving  the  nozzle  be 


244  THERMODYNAMICS 

known.  This  quality  can  be  read  directly  from  the  MoUier 
diagram  at  the  state-point  4,  Fig.  67,  or  it  may  be  com- 
puted by  means  of  the  formulas  developed  in  Exercise  233. 

In  computing  the  area  at  the  throat  of  the  nozzle  it 
is  customary  to  assume  that  the  whole  loss  due  to  friction 
occurs  between  the  throat  and  the  exit  section  of  the  nozzle. 
That  this  assumption  is  reasonable  appears  from  the  facts 
that  the  length  of  the  nozzle  between  entrance  and  throat 
is  very  short  and  that  the  velocity  in  this  part  of  the 
nozzle  is  relatively  low.  Under  these  assumptions  the 
velocity  at  the  throat  is  due  to  the  whole  difference  be- 
tween the  heat  contents  at  the  initial  pressure  and  at  the 
critical  pressure  existing  at  the  throat. 

As  regards  the  length  of  the  nozzle  nothing  can  be 
determined  by  simple  computations.  In  practice  the  rules 
laid  down  on  page  236  are  usually  followed.  For  an  in- 
teresting discussion  on  the  determination  of  the  form  of 
nozzles  see  the  pamphlet  by  Leblanc  referred  to  on  page  230. 

Exercise  235.  Design  a  nozzle  for  the  conditions  given  in 
Exercise  228  assuming  the  ratio  of  the  loss  due  to  friction  in  the 
expanding  portion  of  the  nozzle  to  the  total  energy  available 
under  ideal  frictionless  conditions  to  be  10  per  cent. 

Section  XLII 
THROTTLING 

A  fluid  is  said  to  be  throttled  or  wiredrawn  when  its 
pressure  is  diminished  due  to  its  flow  thru  a  contracted 
passage.  The  flow  thru  pressure-reducing  valves,  ex- 
pansion valves  in  ammonia  refrigerating  machinery,  valves 
of  steam  and  gas  engines  are  cases  in  which  throttling 
occurs. 


FLOW    OF    FLUIDS  245 

The  pressure  on  the  down-stream  side  of  the  constric- 
tion depends  upon  the  nature  and  the  state  of  the  fluid, 
the  amount  of  constriction,  and  the  velocity  of  approach. 
It  is  usually  unnecessary  to  determine  this  pressure.  The 
determination  of  the  drop  in  temperature,  of  the  change 
in  specific  volume,  and  of  the  loss  in  availability  of  the 
energy  of  the  throttled  fluid  for  a  given  drop  in  pressure 
is  however  of  great  importance. 

During  the  throttling  process  illustrated  in  Fig.  68  some 
of  the  potential  energy  of  the  fluid  is  transformed  into 
kinetic   energy   at    the   constriction.     This   kinetic   energy 


Fig.  68. 

is  then  re-transformed,  wholly  or  in  part,  into  heat  by 
internal  friction. 

If  the  flow  is  adiabatic  (altho  not  isentropic)  we  have 

=  JKJLI  — 12) 

H 

as  the  fundamental  equation. 

In  most  cases  the  kinetic  energies  per  pound  of  fluid 
at  sections  i  and  2  are  relatively  small  and  practically 
equal  so  that  their  difference  is  zero.  This  implies  that 
'W2  =  'W\  and  then 

ii=i2. 

Under  these  conditions  the  throttling  process  is  a  curve  of 
constant  heat  content. 


246  THERMODYNAMICS 

Exercise  236.  Show  that  for  ideal  gases 

u  —  CvT,     i  =  CpT, 
also  that  the  temperature  remains  unaltered,  and  that 

piVi  =  P2V2 

during  throttling  for  which  w^  —  Wi. 
Exercise  237.  Show  that  for  ideal  gases 

A1V2 

W2  =  Wi  - — ■ 
A2V1 


and  that 

I     Wi- 
ll—  ^2  =  ";: 

JCp   2g 


.(iS)"-'] 


Exercise  238.  Air  flows  from  a  pipe,  in  which  ^£'1  =  300  feet 
per  second,  into  a  large  receiver.  Compute  the  change  in  tem- 
perature of  the  air. 

Exercise  239.  Show  that  for  superheated  vapor 

1    pv        .1  fnpv 
«  =  — ,     ?=■ 


J m—i  Jm—i 

and  that  for  throttUng  in  which  the  change  in  velocity  may  be 
neglected 

PiVi  =  P2V2. 

Exercise   240.  Assuming  as  the  characteristic  equation  for 
superheated  steam 

pv  =  8s.85T-.2s6p, 

show  by  means  of  this  equation  that  the  throttling  of  superheated 
steam  for  which  7^1  =  ^2  is  always  accompanied  by  a  drop  in 
temperature. 

Von  Linde's  Process  for  the  Liquefaction  of  Gases. — 

An  ideal  gas  suffers  no  change  in  temperature  when  throttling 
occurs  with  wi  =  iV2.  Thomson  and  Joule  showed,  in  1853, 
by  means  of  their  porous  plug  experiments  that  the  throttling 
of  actual  gases  such  as  air,  nitrogen,  oxygen,  and  carbon 


FLOW   OF   FLUIDS  247 

dioxide  produces  a  slight  drop  in  the  temperature  of  these 
gases.  This  slight  cooling  of  the  gases  due  to  throttling 
was  utilized  by  von  Linde  for  the  production  of  liquid 
air  as  follows. 

A  compressor  supplied  compressed  air  at  say  100  at- 
mospheres to  the  inner  tube  of  a  long  double  worm  thru 
which  the  air  passes  to  a  throttling  valve.  This  valve 
reduces  the  pressure  to  say  20  atmospheres  and  the  air 
suffers  a  drop  of  about  27°  F  in  temperature.  The  air 
after  throttling  passes  thru  a  vessel  arranged  to  catch 
the  liquid  air  produced  and  then  returns  thru  the  outer 
tube  of  the  double  worm  to  the  compressor. 

If  the  original  temperature  of  the  air  after  leaving  the 
cooHng  coils  placed  between  the  compressor  and  the  double 
worm  is  say  70°  F,  the  temperature  after  throttling 
will  be  43°  F.  This  colder  air  returning  thru  the 
outer  worm  to  the  compressor  cools  the  air  on  its  way 
to  the  throttling  valve.  The  air  arriving  at  the  valve  soon 
has  a  temperature  of  43°  F,  and  the  throttling  cools  it 
to  16°  F.  After  the  machine  has  been  running  for  several 
hours  the  air  finally  reaches  the  state  of  a  saturated  vapor 
and  Hquid  air  is  deposited  after  throttling.  The  whole 
machine  must  of  course  be  thoroly  lagged  and  a  separate 
compressor  must  transfer  air  from  the  atmosphere  to  the 
high-pressure  cycle. 

The  Throttling  of  Wet  Vapors. — As  already  shown 
ii  =  12  whenever  wi  =  wo. 

Exercise  241.  Show  that  the  quality  of  a  wet  vapor  equals 

Xi  = h-Vi— 

after  throttling  provided  'u!i  =  'u:i. 


248  THERMODYNAMICS 

Exercise  242.  What  percentage  of  moisture  may  steam  at 
150  pounds  per  square  inch  absolute  contain  if  throttling  to 
15  pounds  per  square  inch  absolute  is  to  dry  it? 

Solve  by  means  of  (a)  the  steam  tables,  (b)  the  MoUier  diagram. 

Exercise  243.  Steam  initially  at  150  pounds  per  square  inch 
absolute  is  throttled  to  20  pounds  per  square  inch  absolute. 
The  observed  temperature  after  throttUng  is  248°  F.  Find 
the  initial  quality  by  means  of  (a)  the  steam  tables,  (b)  the 
MoUier  diagram. 

Exercise  244.  Develop  a  formula  for  obtaining  the  initial 
quality  of  steam  by  means  of  the  readings  made  on  a  throttling 
calorimeter  and  by  means  of  the  steam  tables. 

The  Loss  of  Availability  Due  to  Throttling. — On  page 
241,  during  the  discussion  of  the  efifect  of  friction  on  the 
flow  of  fluids,  it  has  already  been  shown  that  an  adiabatic 
is  not  necessarily  an  isentropic  curve. 

During  a  reversible  adiabatic  process  the  fluid  does 
work  during  the  frictionless  expansion  either  upon  a  piston 
or  by  imparting  kinetic  energy  to  its  own  mass.  Under 
these  conditions  the  entropy  of  the  fluid  remains  constant 
and  the  adiabatic  process  is  also  an  isentropic  process. 

During  an  irreversible  adiabatic  process  the  entropy 
increases  even  tho  no  heat  reaches  the  fluid  from  an 
external  source.  Under  these  conditions  some,  or  all,  of 
the  kinetic  energy  developed  during  the  flow  is  transformed 
back  into  heat  and  this  heat  so  returned  to  the  fluid  causes 
the  increase  in  entropy. 

During  throttling  for  which  wi  =  W2  and  during  which 
adiabatic  conditions  exist  no  heat  is  lost  and  yet  the  ability 
of  the  fluid  to  do  work  has  decreased.  No  energy  has  been 
lost  or  gained  but  the  availability  of  this  energy  has  been 
diminished. 


FLOW   or  ^FLUIDS 


249 


To  illustrate,  consider  an  adiabatic  flow  of  an  ideal  gas 
for  which  wi  =  W2  and  therefore  ii  =  i2,  T\  =  T2,  and 
piVi  =  p2V2.  Thus  this  process  altho  adiabatic  is  also  iso- 
thermal as  represented  by  the  lines  i  2  in  Figs.  69  and  70. 
The  gas  has  passed  thru  the  process  i  2  and  yet  the 
area  under  i  2  in  Fig.  69  does  not  represent  the  work  done. 
That  is  even  under  ideal  conditions  no  mechanical  energy 
results  from  this  change  of  state  so  that  no  energy  can 
be  stored  so  as  to  be  available  in  assisting  the  return  of 


Fig.  69. 


the  gas  to  its  initial  condition.  It  might  be  supposed 
that  as  the  whole  heat  energy  in  the  gas  at  i  is  still  in  the 
gas  under  the  conditions  represented  by  2  the  return  of 
the  gas  to  the  condition  i  would  require  no  external  supply 
of  energy. 

That  this  is  not  so  is  evident  when  we  consider  the  effect 
of  compressing  the  gas  iso thermally  from  2  to  i.  The 
external  work  required  to  do  this  is  rejiresented  by  the 
area  under  2  i  in  Fig.  69.  When  this  work  has  been  sup- 
plied the  gas  is  exactly  in  its  original  state  but  some  asso- 


250  THERMODYNAMICS 

dated  body  must  contain  the  heat  rejected  by  the  gas 
during  this  process.  This  heat,  equivalent  to  the  work 
done,  cannot  be  completely  transformed  into  work,  thus 
availability  has  been  lost.  The  work  done  by  and  the 
heat  stored  in  bodies  associated  with  the  gas  during  the 
reversible  isothermal  compression  equals 

-^1.1  log- 

heat  units  per  pound  of  gas.  If  T^  is  the  lowest  obtainable 
temperature  then  only 


(Ti-Tl\i    .  ,        V2 


can  be  again  transformed  into  work  under  even  the  most 

ideal  conditions. 

^Q     I  pivi        V2 
As  S2-si  =  ~  =  -h^log- 

the  loss  of  availability  due  to  the  process  i  2  may  be  ex- 
pressed as  follows, 

=  (S2-Si)Ti, 

the  product  of  the  change  in  the  entropy  of  the  gas  due 
to  the  irreversible  process  and  the  lowest  obtainable  tem- 
perature expressed  in  absolute  degrees. 

It  should  be  noted  that  the  area  under  the  process  i  2 
in  Fig.  70  does  not  represent  heat  supplied  to  the  gas 
during  the  irreversible  adiabatic  process  12.  It  does  rep- 
resent the  heat  equivalent  of  the  work  that  must  be  per- 


FLOW    OF   FLUIDS 


251 


formed  and  of  the  heat  that  must  be  extracted  from  the 
gas  during  its  change  back  to  the  initial  state  by  reversible 
isothermal  compression  from  2  i  along  the  same  process 
line. 

The  gas  may  be  returned  to  its  initial  state  i  by  a 
smaller  expenditure  of  work  than  that  required  by  the 
isothermal  compression  explained  above.  For  instance  let 
the  horizontal  dotted  line  in  Fig.  69  represent  the  lowest 
obtainable  pressure    (the   atmospheric  pressure)   then  the 


Fig.  70. 


gas  may  under  ideal  conditions  be  conceived  to  be  expanded 
along  a  reversible  adiabatic  from  2  to  3,  then  cooled  at 
constant  pressure  from  3  to  4,  and  finally  compressed 
along  a  reversible  adiabatic  from  4  to  i.  The  energy 
expended  during  these  processes  is  less  than  the  work 
required  for  the  isothermal  compression  discussed  above. 

Exercise  245.  Show  that  the  energy  required  to  produce 
the  changes  of  state  indicated  by  the  processes  2341,  Fig. 
6g,  amovuits  to  Jcp{Tz—T^  per  pound  of  gas. 

Exercise  246.  Obtain  the  result  of  Exercise  245  from  Fig.  70, 


252  THERMODYNAMICS 

Even  less  energy  than  this  would  suffice  to  return  the 
gas  from  state  2  to  state  i  if  instead  of  the  lowest  obtain- 
able pressure  use  is  made  of  the  lowest  obtainable  tem- 
perature. As  shown  in  Figs.  69  and  70,  the  gas  at  2  may 
be  expanded  along  a  reversible  adiabatic  to  5,  then  com- 
pressed along  a  reversible  isothermal  at  the  lowest  obtain- 
able temperature  to  4,  and  finally  compressed  adiabatically 
at  constant  entropy  to  i.  The  work  obtained  from  the 
gas  during  the  process  2  5  is  equal  to  the  work  required 
to  compress  the  gas  from  4  to  i.  But  during  the  process 
5  4  work  must  be  done  on  the  gas  by  some  associated  body 
and  heat  equivalent  to  this  work  must  be  rejected  by  the 
gas  to  some  associated  body.  This  heat,  as  may  be  seen 
from  Fig.  70,  equals 

where  Tz,  is  the  lowest  obtainable  temperature,  and  it  is 
wholly  unavailable  for  transformation  into  mechanical 
energy  for  its  temperature  is  already  the  lowest  obtainable. 

The  loss  of  availability  due  to  the  irreversible  adiabatic 
process  equals  the  increase  in  entropy  during  the  irre- 
versible process  multiplied  by  the  lowest  obtainable  tem- 
perature expressed  in  absolute  units.  It  is  still  the  same 
as  the  loss  due  to  the  direct  reversible  isothermal  com- 
pression from  2  to  I. 

During  the  throttling  of  a  vapor  similar  conditions 
exist.  As  shown  in  Fig.  71  in  which  the  points  are  num- 
bered so  as  to  agree  with  Figs.  69  and  70,  the  irreversible 
adiabatic  i  2  coincides  with  a  constant  heat  content  line, 
for  now  i\  =  i2-  The  series  of  processes  by  means  of  which 
the  vapor  at  2  can  be  returned  to  its  initial  condition  i 
with  the  least  expenditure  of  external  energy  is  evidently 


FLOW   OF   FLUIDS 


253 


represented  by  the  line  2541,  provided  the  temperature 
along  5  4  is  the  lowest  obtainable  temperature  Tx,.  Here 
also  the  loss  of  availability  due  to  the  irreversible  process 
I  2  is 

Tl{s2-Si). 


Fig.  71. 

Exercise  247.  Steam  is  supplied  to  an  engine  the  back 
pressure  in  which  is  3  pounds  per  square  inch  absolute.  The 
steam  is  throttled  from  150  pounds  per  square  mch  absolute  and 
a  quaUty  of  0.99  to  1 20  pounds  per  square  inch  absolute  before 
entering  the  engine.  Compute  the  loss  of  available  energy  per 
pound  of  steam  due  to  throttling. 


Section  XLIII 

VENTURI   METERS 

Vcnturi  meters  are  used  for  measuring  the  flow  of  fluids 
in  pipes.  These  meters  are  especially  useful  when  the 
quantity  of  fluid  to  be  measured  is  so  large  that  the  dis- 
placement meters  of  the  usual  sort  capable  of  handling  the 
flow  would  be  very  large  and  correspondingly  costly. 


254 


THERMODYNAMICS 


As  already  described  in  hydraulics  the  Venturi  meter, 
Fig.  72,  consists  simply  of  a  gradual  reduction  in  the  cross- 
section  of  the  pipe  line  and  a  gradual  enlargement  of  the 
section  to  its  original  diameter.  Provision  must  be  made 
for  measuring  the  pressures  at  the  up-stream  section  of  the 
pipe  before  contraction  occurs  and  at  the  throat  of  the 
meter. 

By  means  of  these  pressures,  the  areas  at  the  corre- 
sponding sections  of  the  pipe,  and  the  physical  constants 
of  the  fluid,  the  velocity  and  the  quantity  of  fluid  passing 
any  section  can  be  computed. 


Fig.  72. 


Let  the  subscripts  i  denote  the  state  of  the  fluid  in  the 
up-stream  section  of  the  pipe  and  the  subscripts  2  the 
state  at  the  throat,  then  the  fundamental  equation  is 


2g 


vdp  = (pivi  —  P2V2) 

p.,  in  —  I 


■■ pm    I  -  h- 

w—i        [       \p\ 


where  m  is  the  exponent  of  the  equation  for  reversible 
adiabatic  change  of  state  which  is  assumed  to  occur  between 
I  and  2,  so  that 

p2      \Vlj 


FLOW   OF   FLUIDS 


255 


If  W  represents  the  pounds  of  fluid  passing  any  section 
per  second 

AiWi       A2'W2 


W= 


Vl 


V2 


and 


AiVo  Ai/pi\m 

102  =  Wi—. —  =  'i^i-r\ 

A2V1  A2\p2 


Exercise  248.  Show  that  in  a  V'enturi  meter 


and  that 


W=A, 


For  air  ^=^=1.40  and  vi  may  be  computed  from 
piVi^RTi  by  means  of  the  temperature  of  the  air  taken 
in  the  up-stream  section  of  the  pipe. 

For  steam  the  value  of  W  may  be  computed  more  readily 

and  more  accurately  from  the  fundamental  equations  and 

the  steam  tables  or  diagrams  than  by  means  of  the  formulas 

developed  in  Exercise  248. 

Thus  we  have 

,2. 


W2    —IVr 
2g 


=  J(ii  —  i2) (i) 


where  (11  —  ^2)  can  be  read  from  the  MoUicr  diagram. 


256  THERMODYNAMICS 

AiWi       A2'W2 


Also  as  W=' 


Vi  V2 


A1V2  .   . 

W2  =  Wi- (2) 

A2  Vi  ^    ^ 

from  which,  by  means  of  the  specific  volumes  vi  and  V2, 
W2  can  be  found  in  terms  of  wi.  Substituting  this  value 
in  equation  (i),  we  find  wi  and  finally 

AiWi 


W= 


Vi 


yields  the  number  of  pounds  of  steam  passing  thru  the  pipe. 
Experiments  have   shown   that   the  Venturi   meter   fur- 
nishes a  very  reliable  and  accurate  means  of  measuring 
the  flow  of  fluids. 

Exercise  249.  A  Venturi  meter  has  a  diameter  of  6  inches 
at  entrance  and  3  inches  at  the  throat.  The  steam  entering  this 
meter  contains  one  per  cent  of  moisture  and  is  under  a  pressure 
of  120  pounds  per  square  inch  absolute.  The  pressure  at  the 
throat  is  100  pounds  per  square  inch  absolute.  How  many 
pounds  of  steam  pass  thru  the  meter  per  second? 

Section  XLIV 

FLOW   THRU   PIPES 

The  Hydraulic  Formula. — The  flow  thru  pipes  of  con- 
stant diameter  will  now  be  considered.  As  demonstrated 
in  hydraulics  the  flow  of  liquids  thru  pipes  is  accompa- 
nied by  a  loss  in  pressure  head  and  therefore  of  energy  due 
to  the  frictional  resistances  encountered.  This  loss  of  head  is 

,2 


pi-p2_4fl  v' 
nf= —  —i —  , 


FLOW   OF   FLUIDS  257 

where  /  is  the  friction  factor, 

w  the  specific  weight  of  the  liquid   in  pounds 

per  cubic  foot, 
V   the  velocity  of  flow  in  feet  per  second, 
/  the  length  of  the  pipe  in  feet, 
d  the  diameter  of  the  pipe  in  feet, 
p\  —  p2  the  loss  of  pressure  in  pounds  per  square  foot, 
and  g  =  32.2  feet  per  second  per  second. 

This  formula  expressed  in  the  notation  we  have  used 
in  thermodynamics  becomes 

ftf=(pl-p-2)v  =  ^-, 

where  v  is  the  specific  volume  in  cubic  feet  per  pound 
and     w  is  the  velocity  of  flow  in  feet  per  second. 

When  dealing  with  gases  and  vapors  v  and  w  cannot  be 
regarded  as  constant  because  a  decrease  in  pressure  causes 
an  appreciable  change  in  the  specific  volume  v. 

The  equation  of  continuity  is  now 

W  = =  a  constant, 

V 

and  not  Aw=-d  constant, 

as  is  the  case  when  the  flow  of  liquids  is  considered. 

If  however  the  drop  in  pressure  is  slight  it  may  be  assumed 
that  the  change  in  specific  volume  is  small  and  as  an  approx- 
imation V  may  be  regarded  as  constant. 

The  friction  factor,  /,  is  not  constant.  It  depends  upon 
the  diameter  of  the  pipe  and  the  nature  of  and  the  specific 
volume  of  the  fluid.  Unwin  deduced  from  experiments 
on  the  flow  of  gases 

/=  0.0027  f  I +^j,  for  air. 


258  THERMODYNAMICS 

and  /=  0.0044!  i-\ — ^  j,  for  illuminating  gas, 

where  d  is  the  diameter  of  the  pipe  in  feet. 

Assuming   the   specific  volume  of  the   fluid   to   remain 
constant  we  have 

hf=(pi-p2)v=fj—, 

so  that  Pi  —  p2=f-^ — . 

vd  2g 

It  is  customary    to   express  this   result  in  terms  of   V, 
the  volume  of  fluid  passing  into  the  pipe  per  second. 

A  T/  H"\  4F 

As  V  =  'w\ — I     or    w  =  ^, 

\  4  /  T^d^ 

where     the    velocity   w    is    assumed    to    remain    constant 
thruout  the  length  of  the  pipe, 

we  have  p,-p,  =  -^—^ 


y^     \T^'^gvd^{p\-p2) _^    \vd^{p\-p2) 
\  32//  \  I 

This  is  known  as  D'Arcy's  formula. 

Exercise  250.  The  following  formula  based  upon  D'Arcy's 
formula  is  used  by  engineers  for  all  gases  and  vapors.  When 
pi  —  p2  is  small  it  yields  fairly  accurate  answers. 


where        W  represents  the  pounds,  and 

Vi  the  cubic  feet  of  fluid  passing  into  the  pipe  per 
minute. 


FLOW   OF   FLUIDS  259 

d  is  the  internal  diameter  of  the  pipe  in  inches, 
/  is  the  length  of  the  pipe  in  feet, 
Vi  is  the  specific  volume  of  the  fluid  entering  the  pipe 
in  cubic  feet  per  pound, 
pi  — pi  is  the  drop  in  pressure  in  pounds  per  square  inch. 
What  is  the  value  of  /  used  in  this  formula? 

Exercise  251.  What  weight  cf  dry  saturated  steam  under 
an  initial  pressure  of  120  pounds  per  square  inch  absolute  may 
be  expected  to  pass  thru  a  4-inch  pipe  700  feet  long  when  the 
drop  in  pressure  is  6  pounds  per  square  inch? 

Exercise  252.  If  the  quality  of  steam  flowing  thru  the 
pipe  in  Exercise  251  were  doubled  what  would  be  the  drop 
in  pressure  provided  the  initial  pressure  is  maintained  at  its 
original  value? 

Exercise  253.  Ten  drills  require  iioo  cubic  feet  of  free  air 
(measured  at  14.7  pounds  and  60°  F)  per  minute,  what  should 
be  the  diameter  of  a  pipe  5000  feet  long  supplying  these  drills 
with  compressed  air  at  55  pounds  per  square  inch  gage,  the 
drop  in  pressure  in  the  pipe  line  not  to  exceed  5  pounds  and 
the  air  entering  the  pipe  having  a  temperature  of  60°  F? 
How  fast  does  the  air  flow  thru  this  pipe? 
Exercise  254.  It  is  stated  that  the  approximate  diameter 
of  a  pipe  line  may  be  obtained  from  the  following  formula 


\    (Pi 


ooo^W'Hv 


in  which  the  letters  have  the  same  significance  as  in  Exercise  250. 

(a)  For  what  value  of  d  is  this  formula  correct? 

(b)  Solve  Exercise  253  by  means  of  this  formula. 

A  More  Accurate  Formula  for  the  flow  of  fluids  thru 
pipes  may  be  deduced  by  taking  into  consideration  the 
variation  of  the  specific  volume  and  of  the  velocity  of  the 
flowing  fluid. 


26o  THERMODYNAMICS 

The  fundamental  equations  are  the  equations  of  con- 
tinuity 

TTr        Aw 

W=  — =a  constant, 

V 


and  the  energy  equation  deduced  on  page  217, 

W2^-Wi^  .      ^  fP'^ 

hi/2=-|     vdp, 

where    1/2    represents    the    energy  lost    thru    friction   per 
pound  of  fluid.     This  1/2  equals  hj  so  that 

1/2  =  ///=/ -7  — • 

As  w  is  a  variable  ///  varies.     Assuming  /  constant  for 
a  given  pipe  and  a  given  gas  and  placing 

2gd 

the  head  lost  per  differential  length  of  the  pipe  line  may 
be  expressed  as  follows, 

d(hf)  =  C(dl)u>^. 

Substituting  this  value  into   the  differential  form  of   the 
energy  equation  which  is 

d(i^) 


2g 


-\-d(if2)+vdp  =  o 


we  have  • \-C{dl)w^-irvdp  =  o 

I  d(j^  vdp_ 

or  5 — \-(-dH — 9=0. 

2g    w^  w~ 

The  first  two  terms  of  this  equation  can  be  readily  inte- 


FLOW   OF   FLUIDS  26 1 

grated.     The  third  term  must  first  be  expressed  in  terms  of 
a  single  variable,  preferably  p,  before  it  can  be  integrated. 

The  equation  of  continuity  affords  a  means  of  expressing 
w  in  terms  of  v 

Wi 

w= — V, 

Vi 

for  the  diameter  of  the  pipe  is  assumed  constant. 

To  express  v  in  terms  of  p  let  us  assume  that  the  favorite 

polytropic  relation    of    thermodynamics  pv**=pivi'*,  where 

w    is    a    constant,  holds    during    the    flow  thru    the  pipe. 

Then 

1 
vdp _vi^dp _  v\  ( p 


S> 


and  the  integral  of  the  energy  equation  is 


I  1  9  ,  ^7  1     ^^        ^1 


ra+l 


log,  K'2_f_Q_| . p   „    =(J 

Wi^pin 


2g  "+  I  „,.  2 


To  determine  the  constant  of  integration  Ci,  note  the 
following  conditions  at  the  beginning  and  at  the  end  of 
the  pipe, 

p  —  p\,        w  =  wi,        1=0 

P  =  p2,  1V  =  W2,  l=L, 

where  L  is  the  length  of  the  pipe.     By  means  of  the  con- 
ditions at  the  beginning  of  the  pipe  we  obtain 

C\=—  log,  wi-^H j S-- 

2g  n-\-i  wi" 

The  energy  equation  becomes 

n+l 

I           w^                n    vipi  \  ( p\  ^         ] 
—  loge— 2+C/H — — S-      f-         -i}=o. 


262 


THERMODYNAMICS 


As  we  seek  the  velocity  at  entrance  to  the  pipe  note  that 


1 

•W  _V  _  /pl\n 


and  solve 

iiog.(^)+«a+j^'J4i 

n         \p /  n-f-i  Wi" 

for  wi,  obtaining 


^         -I     =o 


pm    I  - .  . 
«+I       I       \pl 


n+l 


2/L     I  pi 

~-+-  loge  T- 

a      n        p2 


When  L  is  large,  the  term  -  log^  —  may  be  neglected. 

n        p2 

Also  the  temperature  of  the  gas  in  the  pipe  may  be  assumed 
to  remain  constant,  because  any  heat  generated  by  friction 
would  be  dissipated  by  radiation  and  conduction.  There- 
fore in  pv^=a,  constant,  w  may  be  assumed  to  be  unity. 
With  these  assumptions 


Wi 


=V^ 


gdpivi  \  pr—p2 
AfL 


pi' 


or 


Exercise  255.  Show  that 


4^[-m 


Tr^d^g(pi'-p2^) 


64fLRTi 
where  W  is  in  pounds  per  second. 


FLOW   OF   FLUIDS  263 

Exercise  256.  Show  that  the  cubic  feet  of  free  air  (measured 
at  14.7  pounds  and  60°  F)  flowing  per  minute  thru  a  pipe  d 
inches  in  actual  diameter  and  L  feet  long  is 


--Wf!-te)l. 


the  pressures  being  in  pounds  per  square  inch  absolute  and 
the  temperature  of  the  air  entering  the  pipe  being  60°  F. 

Exercise  257.  If  the  drop  in  pressure  in  the  pipe  line  is  small 
show  that  D'Arcy's  formula  may  be  derived  as  an  approximation 
from  the  result  of  Exercise  255. 

Exercise  258.  A  cast  iron  pipe  10.3  miles  long  and  0.98 
foot  in  diameter  is  to  deliver  135  cubic  feet  of  free  air  per  second. 
For  this  pipe  /=  0.0023.  The  initial  pressure  is  92  pounds  per 
square  inch  gage.  Compute  the  terminal  pressure,  the  tem- 
perature thruout  the  pipe  Une  being  60°  F 

(a)  by  means  of  the  more  accurate  method, 

(b)  by  means  of  the  hydraulic  method. 

(c)  Compute  the  initial  and  the  final  velocity  of  flow  in  this 
pipe. 


CHAPTER  XIV 
GENERAL  EQUATIONS  OF  THERMODYNAMICS 

Section  XLV 
DIFFERENTIAL  EXPRESSIONS  FOR  THE  HEAT  SUPPLIED 

The  State  of  a  Body  is  not  determined  by  its  temper- 
ature alone.  It  has  already  been  shown  that  for  ideal  gases 
and  even  for  superheated  vapors  three  quantities  are 
sufficient  to  determine  the  state  of  the  body.  These 
may  be  the  temperature,  the  pressure,  and  the  specific 
volume  and  they  are  always  so  related  by  means  of  a 
characteristic  equation  that  any  two  being  arbitrarily 
assumed  the  third  can  be  found. 

It  is  by  no  means  true  that  these  or  any  other  three 
quantiti;is  always  determine  the  state  of  any  body.  For- 
tunately three  variables  suffice  for  many  cases  and  to 
such  cases  the  following  discussions  and  equations  are 
limited. 

Mathematically  this  condition  is  expressed  when  the 
characteristic  equation  of  the  substance  is 

f{p,v,t)  =  o, 

or  t=^(f){p,v). 

The   Heat    Absorbed   During  any  Change   of   State. 

By  reason  of  the  above  assumption  any  two  of  the 
three  variables  p,  v,  and  /  may  in  general  be  assumed  to  be 
the  independent  variables  during   any  change  of  state  and 

264 


GENERAL   EQUATIONS    OF    THERMODYNAMICS 


265 


the  heat  absorbed  during  this  change  of  state  may  be  com- 
puted in  terms  of  these  variables  and  of  certain  thermal 
capacities. 

Assume  p  and  v  as  the  independent  variables.  Geomet- 
rically this  means  that  the  projection  on  the  /)D-plane  of 
the  actual  change  of  state  which  occurs  on  the  charac- 
teristic surface  (Fig.  2)  is  to  be  considered. 

The  heat  absorbed  during  any  differential  change  of 
state  when  projected  on  the  /?z)-plane  may  be  conceived 


Fig.  73. 

to  occur  in  two  steps  one  while  v  remains  constant,  the 
other  while  p  remains  constant  (Fig.  73). 


Let 


®r"' 


and 


be  the  heat  absorbed  by  a  unit  mass  of 
the  substance  per  unit  change  in  volume 
while  the  pressure  remains  constant, 

(  — )  ^//,.  be  the  heat  absorbed  by  a  unit  mass  of 
the  su])stance  per  unit  change  m  pres- 
sure while  the  volume  remains  constant. 
Then  if  dq  represents   the  whole  heat  absorbed  per  unit 
mass  of  the  substance  during  the  complete  change  of  state 


^=(1);'"+ 


(i-;)/^' 


(l) 


266  THERMODYNAMICS 

because  the  actual  change  in  volume  is  dv  and  the  actual 
change  in  pressure  is  dp. 

The  change  in  temperature  does  not  appear  explicitly 
in  this  equation.  It  is  taken  care  of  by  the  relation 
fip>  v,  t)  =  o  which  means  that  p  and  v  cannot  change  with- 
out a  corresponding  definite  change  in  t. 

dq  is  not  the  heat  absorbed  along  a  differential  element 
of  a  line  on  the  pv-p\a.ne,  but  it  is  the  heat  absorbed  during 
the  corresponding  actual  change  represented  on  the  char- 
acteristic surface. 

Two  other  expressions,  similar  to  equation  (i),  for  the 
same  dq  can  be  deduced  from  the  projections  of  the  actual 
change  of  state  upon  the  other  two  coordinate  planes. 

If  (— ^)  =Cp  represents  the  heat  absorbed  by  a  unit 
mass  of  the  substance  per  unit  change 
in  temperature  while  the  pressure  re- 
mains constant, 

and  ( — -  )  =ip  represents  the  heat  absorbed  by'  a  unit 
mass  of  the  substance  per  unit  change  in 
pressure  while  the  temperature  remains 
constant, 

then  ,,.(|)^,,+  (|),, (.) 

Also  if  {—)  =Cv  represents  the  heat  absorbed  by  a  unit 
mass  of  the  substance  per  unit  change 
in  temperature  while  the  volume  re- 
mains constant, 

and  (^)  —^'>  represents  the  heat  absorbed  by  a  unit 


GENERAL   EQUATIONS    OF   THERMODYNAMICS  267 

mass  of  the  substance  per  unit  change  in  voliime  while 
the   temperature   remains  constant, 

we  have  ,,=  (|)^,,+  (|)  ,, (3) 

It  will  be  noticed  that  Cp  and  Cc  are  by  definition  the 
familiar  specific  heats  at  constant  pressure  and  at  constant 
volume  respectively. 

Ip  and  Iv  are  called  latent  heats  because  the  temperature 
does  not  change  during  the  changes  of  state  to  which  they 
apply. 

hp  and  h„  have  received  no  special  names. 

These  six  coefl&cients  in  the  differential  expressions  for 
the  heat  absorbed  are  called  thermal  capacities. 

In  order  to  use  equations  (i),  (2),  and  (3)  the  relations 
of  the  thermal  capacities  to  each  other  and  to  the  variables 
p,  V,  and  t  must  be  established. 

Relations  Between  the  Thermal  Capacities. — Only 
four  equations  are  available  to  determine  the  six  ther- 
mal capacities.  They  are  the  characteristic  equation  of 
the  substance,  f(p,  v,  t)  =  o,  and  the  three  expressions  for 
dq.  Therefore  two  thermal  capacities  must  be  known 
before  the  others  can  be  computed.  Four  independent 
relations  between  the  thermal  capacities  can  be  established. 

From  equations  (2)  and  (3)  we  obtain  by  elimination  of  dq 

Cpdt-{-lpdp  =  Cvdt-\-lvdv (4) 

This  equation  contains  the  differentials  of  the  three  var- 
iables p,  V,  and  /.  Any  one  of  these  can  be  eliminated 
by  means  of  the  characteristic  equation 

J{p,v,t)  =  o. 


268  THERMODYNAMICS 

Let   us   eliminate   dt.     To    do    this   put  J{p,  v,  t)  =  o    in 
the  form  t=^(j){p,  v),  from  which  we  obtain 

Exercise    259.  Find  dt  for  an    ideal  gas  by  means  of   this 
equation. 

The  above  indicated  elimination  gives 


Altho  the  value  of  dt  depends  upon  the  values  of 
both  dp  and  dv,  dp  and  dv  are  themselves  independent  of 
each  other  (no  definite  line  on  the  characteristic  surface 
having  been  assumed),  dp  and  dv  may  thus  vary  in  any 
manner  with  various  changes  of  state.  Therefore  the  only 
way  in  which  the  last  equation  can  be  true  for  all  values 
of  dp  and  dv  is  under  the  condition  that  the  coefficients 
of  dp  and  dv  be  both  separately  equal  to  zero. 

Thus  icp-Cp)  =  lvl~)  , (5) 

and  ^^^■"'^''^^~^^(^) ^^^ 

Other  relations  between  the  thermal  capacities  may  be 
found  by  eliminating  v  from  equation  (4)  by  means  of 

obtained  from  f(p,  v,  t)  =  o. 


GENERAL   EQUATIONS    OF    THERMODYNAMICS  269 

In  this  way  the  relations 


and  ^'='\'^p 

are  established. 

The  first  one  of  these  has  already  been  found,  equation 
(5).  The  second  one  altho  apparently  a  new  relation  is 
really  a  result  which  may  be  obtained  from  equations  (5) 
and  (6)  as  follows. 

From  the  calculus  we  have 

when  2=/(-v,  y), 

9.1-         9z  ■ 

ay 

Therefore  from  the  equation 

t  =  ci>{p,v) 

di 

dp_      dv 

dp 
From  equations  (5)  and  (6) 

or  h=  — TT^e 

op 

dt 

and  Ip^TrrU. 

dp 


2'JO  THERMODYNAMICS 

By  the  elimination  of  p  from  equation   (4),  we  obtain 
the  relations 


and 


'•-'(f).' 


both  of  which  have  already  been  obtained. 

Thus  so  far  we  have  found  two  relations  for  the  deter- 
mination of  four  thermal  capacities. 

Exercise  260.  Assuming  Cj,  and  k  to  be  known  for  a  certain 
ideal  gas,  compute  the  latent  heats  of  this  gas. 

Starting  with  equations  (i)  and  (2),  we  have 

lipdv-\-livdp  =  Cpdt-\-lpdp (7) 

Eliminating  dv  the  relations 

and  '•='"{^), 

are  obtained. 

Exercise  261.  Deduce  the  relations 

dt 


to — ftp  —  —  Cb  I 

and  ''M§); 


GENERAL   EQUATIONS   OF   THERMODYNAMICS  27 1 

The  nine  relations  between  the  thermal  capacities  de- 
duced above  are  not  independent  relations.  As  there 
exist  only  four  equations  which  can  be  used  to  determine 
these  relations  only  four  independent  relations  can  be 
found.  Any  four  of  the  above  nine  relations  involving 
the  six  thermal  capacities  will  serve  to  compute  the  six 
thermal  capacities  when  two  of  them  are  known. 

Section  XL VI 

EXACT    DIFFERENTIALS 

Definition. — Many  of  the  differential  equations  of  ther- 
modynamics are  of  the  form 

dz  =  Mdx-\-Ndy, 

where  M  and  N  may  be  constants  or  functions  of  the  var- 
iables X  and  y. 

As  an  example  we  have 

dt=(^)  dv+C^dp,       ....     (a) 


v9^7p         \dp 
which  is  obtained  by  differentiating 

t=4>{p,v). 

Here  M=(^      and    N 

\dv/p 

As  another  example  consider 


"^-(BMU)."' ^« 


where    M=(^]=hp    and  N=(^)=/i^ 
\dv/p  \dp/v 


272  THERMODYNAMICS 

Expressions  of  the  form  Mdx-\-Ndy  are  divided  into 
two  classes.  For  every  expression  belonging  to  the  first 
class,  of  which  the  right-hand  member  of  equation  (a)  is 
an  example,  a  definite  relation  exists  between  the  variables, 
in  the  example  the  relation  is  t=(l>{p,v).  For  all  expres- 
sions belonging  to  the  second  class,  of  which  the  right- 
hand  member  of  equation  {b)  is  an  example,  no  such  func- 
tional relation  exists. 

Whenever  a  functional  relation  between  the  variables 
exists  the  value  of  2  depends  only  upon  the  initial  and  the 
final  values  of  the  independent  variables  x  and  y  and  not 
in  any  way  upon  the  manner  in  which  the  change  is  made 
from  the  initial  to  the  final  condition. 

Referring  to  the  first  example,  note  that  its  integral 
t=(f)(^p^i,)  definitely  fixes  the  change  in  the  temperature 
when  the  values  of  p  and  v  are  known  for  both  the  initial 
and  the  final  conditions  without  any  reference  to  the  path 
the  state-point  may  have  followed  on  the  characteristic 
surface,  as  given  by  some  relation  between  p  and  v,  during 
its  motion  from  the  initial  to  the  final  point. 

When  these  conditions  are  satisfied  Mdx-\-Ndy  is  called 
an  exact  differential. 

Under  these  conditions  the  integration  of 

can  be  performed  without  using  a  relation  between  p  and 
V  and 

I    dt=t2  —  h, 

the  change  in  temperature,  is  always  the  same  when  con- 
puted  between  the  same  initial  and  final  points,  {pi,  Vi) 


GENERAL   EQUATIONS   OF   THERMODYNAMICS  273 

and  {p2,  V2),  irrespective  of  the  line  on  the  surface,  t  =  <l>(p,  v), 
joining  these  points  along  which  the  integration  may  be 
performed. 

Exercise  262.  Write  equation  (a)  so  that  it  applies  to  an  ideal 
gas  and  from  it  determine  the  change  in  temperature  due  to 
a  change  from  (pi,  Vi)  to  {p2,  V2). 

If  no  functional  relation  exists  between  the  variables 
X,  y,  and  2  in 

dz=Mdx+Ndy 

then  the  integration  cannot  be  performed.  No  value  of 
2  can  be  found  unless  some  assumption  regarding  the  re- 
lation between  x  and  y  is  made  so  that  Mdx-\-Ndy  can 
be  expressed  in  terms  of  one  variable.  If  some  relation 
between  .v  and  )'  is  arbitrarily  assumed  each  assumption 
will  yield  a  different  value  of  2. 

Referring  specifically  to  equation  (b),  the  second  exaniple 
above,  we  know  from  physical  considerations  that  the  value 
of  q2—qi  is  not  a  definite  constant  quantity  depending 
only  upon  the  initial  and  the  final  values  of  p  and  v.  The 
heat  supplied  between  any  two  states  of  a  substance  de- 
pends not  only  upon  these  states  but  also  upon  the 
manner  in  which  the  heat  is  supplied.     Therefore 

is  not  an  exact  differential. 


MUh 


F^XERCiSE  263.  Find  Zi—Zi  between  the  limits  (x=o,   v  =  o 
and  {x=i,  y=i)  when 
(i)  dz  =  xdy  —  ydx, 

(2)  dz  =  xdy+ydx, 

provided  (a)  y  =  2iX,  {b)  y-  =  gx. 


274  ■      THERMODYNAMICS 

Which  of  these  expressions  is  an  exact  differential?  Which 
can  be  integrated  directly? 

The  Test  for  an  Exact  Differential.— The  following 
criterion  which  is  used  to  test  expressions  of  the  form 
Mdx-\-Ndy  for  exactness  may  also  be  used  in  another  way. 
Suppose  an  expression  in  this  form  is  known  to  be  exact, 
let  us  say  from  physical  considerations,  then  the  test  about 
to  be  developed  may  be  used  to  establish  relations  between 
the  quantities  involved  in  this  expression. 

If  Mdx-\-Ndy  is  exact  then,  by  definition,  some  function 
z=f(x,  y)  must  exist  which  on  differentiation  gives 

dz  =  Mdx-\-Ndy. 
From  z=f{x,  y) 


we  have  dz  = 


=(i)-+(l)> 


so  that  M=|^    and    N=^. 

dx  dy 

Differentiating   these  identities   the   first   with   respect    to 
y  and  the  second  with  respect  to  x  we  have 

and 


dy      dxdy  dx     9y  ax- 

But  as  has  been  shown  in  the  calculus 

dxdy    dydx' 

so  that  ~^^=^l~- 

dy      dx 


GENERAL   EQUATIONS   OF   THERMODYNAMICS  275 

That  is,  if  the  change  in  a  function  z  of  two  independent 
variables  x  and  y,  between  fixed  limits,  for  which 

dz  =  Mdx-\r^^y 

is  independent  of  the  manner  in  which  the  change  is  pro- 
duced, i.e.,  if  Mdx-\-Ndy  is  an  exact  differential,  then 

av     ax* 

Exercise  264.  The  following  differential  must  be  exact, 

{x^^S^y^y)dx^{y^^Ax--Vx)dy. 
What  is  the  value  of  ^4? 

Section  XLVII 
THE  DIFFERENTIAL  EQUATIONS    OF  THERMODYNAMICS 

Entropy  As  a  Coordinate. — The  coordinates  we  have 
already  used  in  defining  the  state  of  a  substance  are  the 
pressure,  the  specific  volume,  and  the  temperature  (/?,  i>, 
and  /). 

To  these  may  be  added  the  entropy,  s,  which  for  re- 
versible processes  is  defined  by 

dq 

where  T  represents  absolute  temperature,  so  that 

r=/-l-a  constant. 

It  is  important  to  show  that  the  change  in  entropy 
dcDends  only  upon  the  change  of  state  which  occurs  and  not 


276 


THERMODYNAMICS 


in  any  way  upon  the  manner  in  which  the  change  is  made. 
In  passing  from  one  state  to  another  the  change  in  the 
entropy  of  the  substance  will  be  the  same  no  matter  what 
reversible  path  is  followed  in  passing  from  the  initial  to 
the  final  state.     That  this  is  so  may  be  shown  as  follows. 

Any  cycle  may  be  divided  into  differential  elementary 
Carnot  cycles  as  shown  in  Fig.  74.  The  sum  of  these 
elementary  cycles  will  be  the  same  as  the  given  cycle. 


Fig.  74 

For  each  of  these  elementary  cycles  we  have 
dqi^dqz 

dqi     dq2_ 
Ti      T2~°' 

where  dqi  and  dq2  are  the  quantities  of  heat  absorbed 
and  rejected  by  the  working  substance  during  the  cycle 
at  the  absolute  temperatures  Ti  and  T2  respectively,  see 
page  133.  t 


or 


GENERAL   EQUATIONS    OF    THERMODYNAMICS 


277 


Therefore  in  any  reversible  cycle  for  any  substance 


r 


T 


--0, 


or  the  sum  of  the  changes  in  entropy  of  any  substance 
undergoing  a  reversible  cycle  of  changes  must  be  zero. 

Now  consider  the  change  in  entropy  of  a  substance 
during  any  change  of  state  represented  by  the  line  i  .1  2, 
Fig.  75.     It  would  at  first  sight  appear  that  as    the    heat 


P 


Fig.  75- 

supplied  during  the  change  of  state  varies  with  the  nature 

X^dq 
-j;  would  also  vary  with 

the  path. 

That  this  is  not  so  becomes  evident  when  we  consider 
the  return  from  state  2  to  state  i  by  any  other  path  such 

as  2  .S  I.     As    I    -f=o,  the   gain  in  entropy  from  i  to  2 

along  the  path  i  A  2  must  equal  the  loss  in  entropy  from 
2  to  I  along  the  path  2  B  i,  or 

I  —  along  path  i  A  2  =  J  ^  along  the  path  1^2. 


278  THERMODYNAMICS 

Therefore  5  depends  only  upon  the  state  of  the  sub- 
stance and  not  upon  the  manner  in  which  the  state  was 
reached.    Also 


An 

and  ds=-;^  must  be  an  exact  differential. 

Of  the  four  coordinates  p,  v,  t,  and  s  which  are  functions 
of  the  state  only  and  do  not  depend  upon  how  the  state 
may  have  been  attained  any  two  may  be  used  as  inde- 
pendent variables  to  determine  the  state  of  the  substance. 
Two  of  these  variables  are  alwavs  sufficient  for  this  pur- 
pose; by  means  of  these  the  other  two  can  always  be 
computed. 

Thermodynamic  Potentials.  —  The  internal  or  in- 
trinsic energy,  u,  of  unit  mass  of  a  substance  under  cer- 
tain conditions  may  be  regarded  as  the  energy  in  the 
substance  due  to  its  present  state.  It  may  also  be  regarded 
as  the  ability  the  substance  possesses  to  perform  work  due 
to  its  thermal  condition.  From  this  point  of  view  u  may  be 
called  a  thermal  potential;  m  is  a  measure  of  the  potency 
of  the  substance. 

Evidently  u  depends  only  upon  the  state  of  the  sub- 
stance and  not  upon  how  the  state  may  have  been  attained, 
therefore  du  is  an  exact  differential. 


dq  =  du-\-—pdv 


du=dq—jpdv. 


GENERAL   EQUATIONS    OF   THERMODYNAMICS  279 

Thus    dq — jpdv  is  also  an  exact  differential,  altho  neither 

dq  nor  pdv  are  exact. 

The  five  quantities,  p,  v,  T,  s,  and  n  are  all  functions  of 
the  state  only  therefore  any  combinations  of  these  quan- 
tities must  also  be  functions  of  the  state  only.  Three 
such  combinations  have  been  found  very  useful  in  ther- 
modynamic discussions.  They  are  denoted  by  the  letters 
i,  f,  and  4>  and  are  defined  by  the  identities 

i  =  u-|-jpv, 

f  =  u-Ts, 
0  =  i-Ts. 

u,  i,  f,  and  0  are  called  thermodynamic  potentials.  -  Of 
these  u  and  i  are  the  most  useful  to  the  engineer.  They 
have  received  special  names,  u  is  the  internal  energy 
whose  physical  significance  is  well  known,  and  i  is  called 
the  heat  content  for  the  physical  significance  of  which 
see  page  166. 

The  other  two  potentials  /  and  4>  have  received  no  spe- 
cial names;  they  are  simply  known  as  thermodynamic 
potentials.  Applications  of  these  potentials  are  to  be 
found  in  physics  and  in  chemistry. 

The  differentials  of  the  four  thermodynamic  potentials 
u,  i,  f,  and  0  are 

du  =  dq——pdv, 
di  =  du-\-—pdv-\—jvdp, 


28o  THERMODYNAMICS 

dj=  du—  Tds—sdT, 
and  dcf)  =  di  —  Tds —sdT. 

These  may  all  be  expressed  in  terms  of  the  four  coor- 
dinates p,  V,  T,  and  5  and  each  one  in  terms  of  two  of  them 
as  independent  variables.  , 

For  instance,  as  dq  =  Tds 

du  =  Tds-^pdv (8) 

Using  this  relation  we  may  transform  dl  into 
di=  I  Tds  —  —pdv  \  -\-—pdv-\--jvdp 

or  di=Tds+^vdp (9) 

Exercise  265.  Show  that 

df=  —sdT — -pdv,       (lo) 

and  d(f)  =  -vdp  —  sdT (n) 

Note  that  as  u,  i,  f,  and  4>  depend  only  upon  the  state  of  the 
substance  their  differentials  are  exact. 

Maxwell's  Relations. — The  following  four  relations 
between  the  variables  p,  v,  T,  s,  which  are  used  as  coor- 
dinates for  the  determination  of  the  state  of  a  substance 
in  thermodynamics  may  be  readily  derived  from  the  four 
differentials  of  the  potentials.  These  relations,  known 
as  Maxwell's  relations,  are  used  in  establishing  relations 
between  the  thermal  capacities  and  the  variables  p,  v, 
and  T. 


GENERAL   EQUATIONS    OF   THERMODYNAMICS  28 1 

From  equation   (8),  if  we  remember  that  the  differential 
is  exact,  we  have 


\d(n 


ds 


.l^U-ifl-^), (.) 


dv/s~     J\ds 
Exercise  266.  Show  that 

'dT\  ^i/dv_ 
,dp/s     J\ds 


(13) 


Wr^A^).' ^''^ 


dp/T~     J\dT 


(is) 


These  are  Maxwell's  relations.  They  hold  for  all  revers- 
ible changes  of  state  and  for  all  substances  whose  char- 
acteristic equation  is  f(p,v,T)  =  o.  Note  the  physical 
meaning  of  these  equations.  For  instance  from  equation 
(13)  we  see  that  the  rate  of  change  of  the  temperature  with 
respect  to  the  change  in  pressure  along  an  isentropic  path 

must  always  equal  —  times  the  rate  of  change  of  the  vol- 
ume with  respect  to  the  change  in  entropy  along  an  iso- 
piestic  path  for  any  substance  at  any  particular  state. 

Exercise  267.  WTiat  geometrical  interpretation  may  be 
attached  to  equation  (15)? 

Equation  (12)  may  be  written  in  terms  of  q  instead  of  s. 
This  will  show  its  connection  with  the  thermal  capacities 
defined  on  pages  265-267.    As  dq=  Tds  we  have 


\dv)a        AdqJi 


282  THERMODYNAMICS 

Note  that  in   { —  )     q  now  takes    the  place  of  5  as  the 

\dv/g 

variable  which  remains  constant  during  the  change  con- 
sidered. 

But  as  by  definition  ( :r^ )  =  /?p 

we  may  write  /?o=  —  —  I  —  I  . 

Also  from  equation  (3), 

dq  =  Cvdt-\-kdv; 

now  as  q  is  constant,  dq  =  o,  and 

'd'v\  ^_^ 


and                                        ^^  =  7(^1' 
Exercise  268.  Show  that 

.     .     .     (16) 

^'~Up)t~  AdTj,'  •  • 

.     .     .     (17) 

A  General  Expression  for  Cp—Cv — As  an  application 
of  Maxwell's  relations  let  us  find  an  expression  for  (cp—Cv) 
in  terms  of  p,  v,  and  T. 

From  equation  (5) 


Cp      Co  —  /ji  1  I    , 

Iv  may  be  eliminated  by  means  of  equation  (16).    Equation 
(16)  may  be  written 

for  T  =  t-\-a,  constant. 

(0,-0=^(1)^(1)^ ('«) 


GENERAL   EQUATIONS    OF   THERMODYNAMICS  283 

This  relation  is  true  for  all  substances  having  a  characteristic 
equation  of  the  form/(^,  v,  /)  =  o. 

Exercise  269.  Show  that  altho  cp  and  Cv  vary  Cp—Cv  must 
be  constant  for  all  substances  whose  characteristic  equation  is 
pv  =  RT. 

Exercise  270.  Find  Cp  —  c^  for  superheated  steam  assuming 
that  Zeuner's  characteristic  equation 

pv^RT-Cp"*, 
where  R,  C,  and  n  are  constants,  is  correct. 

The  Heat  Supplied,  dq. — The  heat  supplied  during 
any  change  of  state  to  any  substance  whose  characteristic 
equation  is  f{p,  v,  t)  =  o  or  t=4>{p,  v)  can  now  be  expressed 
in  terms  of  Cp,  Cv,  and  the  partial  derivatives  of  p  and  of  v 
with  respect  to  /.  To  do  this  the  thermal  capacities  h, 
Ip,  hv,  and  hp  must  be  eliminated  from  equations  (i),  (2), 
and  (3)  by  means  of  the  relations  just  deduced  from  Max- 
well's relations. 

From  equation  (3)  by  means  of  equation  (16) 

dq  =  c„dt+^(^)^dv,      ....     (19) 

and  from  equation  (2)  by  means  of  equation  (17) 

dq  =  c.dt-^(|^)^dp (20) 

Exercise  271.  From  equations  (19)  and  (20)  deri\e  expres- 
sions for  dq  for  ideal  gases. 

To  simplify  equation  (i),  i.e., 


HfX^-HM)/"' 


284  THERMODYNAMICS 

note  that  I  ;r- )   =lip  =  lv-\-Cvi ;—]  ,  Exercise  261, 

that  ^"—^{^i  J  equation  (16), 

and  that  (:ti)  =fh  —  Cvi-:—)  ,  Exercise  261. 

\dp/v  \dp/v 

Substituting  these  values  in  equation  (i) 

—-]     and  (t—)    use  equation  (18) 
dv/p  \dp/v 

from  which 

T/dp\ 

'dt\      J\dt)v 


dv/p     (cp—cv)' 


T 


and  ,d'\_Adl), 


,dp/v       {Cp—CvY 

Finally  equation  (i)  becomes 

Exercise  272.  What  form  does  equation  (21)  take  for  ideal 
gases?   Compare  the  result  with  the  third  equation  in  Exercise  26. 

Clapeyron's  Fomuila. — Let  us  apply  equation  (19),  i.e., 

which  is  true  for  any  substance  and  which  has  already 
been  applied  to  an  ideal  gas  in  Exercise  271,  to  a  vapor 


GENERAL   EQUATIONS    OF   THERMODYNAMICS  285 

during  the  process  representing  the  change  from  a  liquid 
to  a  saturated  vapor. 
For  a  wet  vapor  we  have 

p=m 

and  v  =  (f){p,x), 

where  x  is  the  quality  of  the  vapor,  see  page  133. 

Therefore   \^]   is  simply  -77-  and  as 
V9/  /v  ^  -^  (It 

dp  . 

-7^  IS  a  constant  for  any  temperature. 

dl' 

If  the  vapor  forms  at  constant  pressure  the  temperature 
is  constant  and  dt  is  zero  so  that 

As  q"  —  q'  =  r  =  ihe  latent  heat  of  vaporization 

rj      I 

we  have  ^" ~" '^' ^  T ' Td^Y ^^^^ 

U) 

This  is  Clapeyron's  formula  for  the  increase  of  volume 
during  the  evaporation  of  a  liquid.  From  it  when  r  and 
the  relation  between  p  and  /  have  been  determined  experi- 
mentally v"  —  v'  can  be  computed. 

Exercise  273.  Deduce  Clapeyron's  formula  directly  from 
Maxwell's  relation  given  in  equation  (14). 


286 


THERMODYNAMICS 


Other  Important  Relations. — The    change    in    internal 
energy 

du=dq——pdv 

may  with  the  aid  of  equation  (19)  be  expressed  in  terms 
of  Cv,  thus 


/9c A  ^2 
\dv)t    J- 


rential 


and  as  du  is  an  exact  differential 

""air 


dt 


or 


which  shows  the  variation  of  Cv  with  respect  to  v  while  the 
temperature  remains  constant. 
Similarly  in 

di  =  dq-\-—vdp,  equation  (9), 

Cp  may  be  introduced  by  means  of  equation  (20),  and  we 
obtain 


M=c^,-^AT{^)-.\ip 


Exercise  274.  Show  from  this  equation  that 

9Cp\  _     T/d-v\ 
dp/t~     J\dr-)v    ' 

which  is"  called  the  relation  of  Clausius. 


(24) 


GENERAL   EQUATIONS   OF   THERMODYNAMICS  287 

Exercise  275.  Show  by  means  of  equations  (23)  and  (24)  that 
for  ideal  gases  Cv  does  not  depend  upon  the  volume,  that  Cp 
does  not  depend  upon  the  pressure,  and  that  nevertheless  both 
Cp  and  Ce  probably  vary  with  the  temperature. 

From  the  equation 


di  =  c.dt-jlT{gj-v\dp 


we  find  that  during  an  isothermal  process  for  which  dt 
must  equal  zero 

l)r-7{<i)r+  •  •  •  <-' 

Also  during  throttling,  for  which  i  remains  constant  so 
that  di  =  o,  such  as  occurs  in  the  Joule-Thomson  porous 
plug  experiment  the  same  equation  yields 


\dp/i    CpJ  ]      \di/ p 

(  — ]  ,  the  ratio  of  the  drop  in  temperature  to  the  droj) 

in  pressure  during  a  process  for  which  the  heat  content 
is  constant,  is  called  the  Joule-Thomson  coefficient. 

Exercise  276.  What  drop  in  temperature  occurs  during  the 
throttling  of  an  ideal  gas? 

The  equality  of  the  absolute  temperature  used  in  the 
general  equations  of  thermodynamics  and  the  absolute 
temperature  in  the  characteristic  equation  of  an  ideal  gas 
may  be  shown  by  means  of  the  equation 


du=c^t^j\  tI^^- p\dv. 


288  THERMODYNAMICS 

Note  that  in  this  equation  du,  the  change  in  internal  energy, 
is  independent  of  the  change  in  volume  when  this  equa- 
tion is  appKed  to  an  ideal  gas,  therefore  the  coefficient 
of  dv  must  be  zero,  or 

(l)rl- 


But  from  pv  =  RT 

(dp\  ^R^p 

therefore  the  two  T's  must  be  alike. 


PROBLEMS  FOR  REVIEW 


277.  Determine  the  position  of  the  absolute  zero  on  the 
Fahrenheit  scale  from  the  following  experimental  result  found 
by  heating  air  at  constant  volume, 

pressure  at  100°  C 

,     o  r^   =  1-3665. 

pressure  at  o    C 

278.  Compute  the  internal  energy  of  a  mass  m  of  an  ideal 
gas  whose  state  is  p,  V,  T  in  terms  of  p,  V,  and  k. 

279.  State  four  laws  governing  the  behavior  of  an  ideal  gas. 

280.  Deduce  the  law  pV  =  mRT  from  the  laws  of  IJoylc  and 
of  Charles. 

281.  Fifty  pounds  of  air  enclosed  in  a  receiver  at  a  pressure  of 
200  pounds  per  square  inch  gage  and  at  210°  F  are  cooled  by 
radiation.  What  will  be  the  final  pressure  if  the  temperature 
of  the  air  becomes  50°  F? 

282.  If  one  cubic  foot  of  o.xygen  at  n.t.p.  weighs  w  pounds, 
compute  the  gas  constant,  R,  for  a  gas  whose  molecular  weight 
is  tx. 

283.  At  a  pressure  of  14.7  pounds  per  square  inch  absolute 
and  a  temperature  of  32°  F  one  pound  of  air  occupies  a  volume 
of  12.39  cubic  feet.  The  specific  heats  of  air  arc  said  to  be 
Cp  =  o.2375  and  (:o  =  o.i685.  From  these  data  compute  the  gas 
constant  for  air  by  two  methods. 

284.  Deduce  a  formula  for  the  gas  constant  of  a  mixture 
of  three  ideal  gases  the  gas  constants  of  which  are  known. 

285.  A   receiver    (capacity    5   cubic   feet)    contains   9   pounds 

280 


2gO  THERMODYNAMICS 

of  SO2.  At  what  temperature  will  the  pressure  in  the  tank 
reach  200  pounds  per  square  inch  gage?  Assume  that  one 
pound  of  oxygen  occupies  11.2  cubic  feet  at  n.t.p.  and  that 
SO2  behaves  as  an  ideal  gas  under  the  above  conditions. 

286.  Deduce  the  relation  between  Cp  and  Cv  of  an  ideal  gas 
and  state  the  law  (or  laws)  governing  the  behavior  of  ideal  gases 
used  in  deducing  this  relation. 

287.  Develop  from  fundamental  equations  the  relation  between 
the  pressures  and  the  temperatures  of  two  states  of  a  given  mass 
of  gas  during  an  adiabatic  expansion. 

288.  Prove  that 

_R     k 

289.  How  much  external  work  is  performed  by  a  gas  during 
(a)  an  adiabatic,  {b)  an  isothermal,  (c)  a  polytropic  change  from 
the  state  determined  by  pi  and  Vi  to  the  state  determined  by 
pi  and  V2. 

290.  During  the  compression  of  one  pound  of  air  10  B.t.u. 
are  removed  by  the  cooling  water  and  30,000  foot-pounds  of 
work  are  required.  The  initial  temperature  of  this  air  was 
80°  F,  compute  the  final  temperature. 

291.  The  temperature  of  air  expanding  in  an  air  motor  drops 
120°  F.  The  expansion  follows  the  process  pv^-^  =  c.  How  much 
heat  was  transferred  to  or  from  the  cylinder  walls  per  pound 
of  gas  during  this  expansion? 

292.  Compute  the  least  amount  of  work  required  per  pound 
of  dry  air  to  draw  it  from  the  atmosphere  at  70°  F  and  deliver 
it  to  a  receiver  at  100  pounds  per  sci''are  inch  gage  under  ideal 
conditions.     The  available  cooling  water  has  a  temperature  of 

7o°F(i?=53-3)- 

293.  Define  polytropic  change  of  state. 

What  is  the  value  of  the  specific  heat  of  an  ideal  gas  during 
polytropic  change  of  state? 


PROBLEMS    FOR   REVIEW  291 

Do  the  various  polytropic  changes  include  all  possible  changes 
of  state? 

294.  How  would  you  determine  whether  or  not  a  given  expan- 
sion or  compression  line  on  an  indicator  card  represents  a  poly- 
tropic change  of  state? 

295.  The  equation  of  the  compression  line  of  an  air  com- 
pressor card  is  found  to  be  pV^-^^  =  a  constant.  During  the 
compression  the  temperature  of  the  air  is  raised  200°  F.  How 
much  heat  is  transmitted  to  the  cylinder  walls  during  the  com- 
pression of  one  pound  of  gas? 

296.  During  a  compression  following  the  law  />T^>-^  =  constant 
the  temperature  of  the  air  rises  from  80  to  200°  F.  How 
much  heat  has  been  absorbed  per  pound  of  air? 

297.  How  much  external  work  was  done  by  each  pound  of 
air  during  the  change  of  state  described  in  296? 

298.  A  balloon,  capacity  9000  cubic  feet,  is  filled  with  air  at 
300°  F.  The  temperature  of  the  surrounding  air  is  70°  F.  What 
total  weight  (including  the  weight  of  the  balloon)  is  required 
to  prevent  its  rising  in  still  air? 

299.  Deduce  a  relation  between  T  and  v  for  an  adiabatic 
change  of  state  of  a  gas  for  which  Cv  =  a+bT,  the  result  to  con- 
tain the  constants  a,  b,  R,  and  /. 

300.  The  state  of  one  pound  of  a  certain  gas  for  which  ^=1.4 
is  changed  from  15  pounds  per  square  inch  absolute,  80°  F. 
and  192  cubic  feet  to  30  pounds  per  square  inch  absolute  and 
250°  F.     Compute  the  change  in  internal  energy. 

301.  During  the  compression  of  an  ideal  gas  40,000  foot- 
pounds of  work  are  expended  and  8  B.t.u.  are  taken  from  it 
by  conduction.  What  change  in  internal  energy  occurs  during 
this  process? 

302.  A  tank  held  40  cubic  feet  of  air  at  200  pounds  per  square 
inch  gage.  This  air  is  cooled  in  the  tank  to  70°  F  and  its  pres- 
sure is  now  1 50  pounds  per  square  inch  gage. 

(a)  How  much  heat  has  been  extracted? 


292  THERMODYNAMICS 

(b)  How  much  has  the  entropy  of  this  gas  decreased? 

303.  Compute  the  change  in  the  entropy  of  an  ideal  gas  due 
to  a  change  of  state  from  (pi,  Vi,  Ti)  to  {p2,  V2,  T2)  in  terms  of 
the  temperatures  and  the  volumes. 

304.  What  is  the  temperature  in  the  exhaust  pipe  of  an  air 
engine  if  the  air  is  supplied  at  100  pounds  per  square  inch  gage 
and  70°  F  and  the  expansion  is  assumed  to  be  adiabatic  to 
atmospheric  pressure? 

305.  Three  cubic  feet  of  air  at  atmospheric  pressure  are  to 
be  compressed  at  constant  temperature  to  100  pounds  per  square 
inch  gage. 

(a)  How  much  heat  must  be  withdrawn  from  the  gas  to  ac- 
complish this? 

(6)  Where  does  this  heat  come  from? 

306.  (a)  Sketch  the  /»i)-diagram  showing  the  operation  of  an 
air  compressor  without  clearance  in  which  the  compression  is 
completed  in  two  adiabatic  stages. 

{b)  Sketch  upon  the  7'5-plane  that  portion  of  the  diagram 
drawn  in  (o)  which  shows  the  saving  effected  by  compounding. 

307.  (a)  Indicate  on  the  />ZJ-plane  an  area,  finite  in  all  its 
dimensions,  which  is  proportional  to  the  heat  supplied  during 
a  change  of  state  from  {pu  Vi,  Ti)  to  {pi,  V2,  T2).  Prove  that 
this  area  represents  the  heat  supplied. 

(b)  Indicate  an  area  on  the  T^-plane  which  represents  the  heat 
supplied  during  the  above  change  of  state. 

308.  In  an  air-compressor  10  cubic  feet  of  air  at  a  temper- 
ature of  100°  F  and  under  a  pressure  of  14  pounds  per  square 
inch  absolute  are  compressed  adiabatically.  The  final  pressure 
is  60  pounds  per  square  inch  absolute,  (a)  Compute  the  volume 
of  this  air  after  compression,  (b)  Find  the  temperature  of 
this  air  after  compression. 

309.  Compute  the  most  advantageous  intermediate  pressures 
for  a  three-stage  compressor  with  perfect  intercooling. 

310.  Compressed   air  is   to   be   stored   in   cylinders   6   inches 


PROBLEMS    FOR    REVIEW  293 

in  internal  diameter  and  6  feet  long.  These  cylinders  can  with- 
stand an  internal  pressure  of  2000  pounds  per  square  inch 
absolute.  The  greatest  temperature  to  which  they  are  liable 
to  be  exposed  is  120°  F.  (a)  What  weight  of  air  can  be  safely 
stored  in  each  cylinder?  (b)  What  should  be  the  charging  pres- 
sure at  60°  F? 

311.  Air  is  compressed  adiabatically  from  pi  and  Ti  and 
delivered  at  p^.  How  much  work  must  be  suppUed  per  pound 
of  air?  Assume  ideal  conditions  and  no  clearance.  Express 
the  result  in  terms  of  the  given  pressures  and  temperature. 

312.  Compressed  air  is  admitted  to  the  cyUnder  of  an  air 
motor  without  clearance.  Compare  the  work  done  during 
admission  with  the  work  it  could  perform  if  expanded  indefi- 
nitely according  to  the  law  pv^  =  c  after  admission  to  the  cyUnder. 

313.  An  air-compressor  has  a  displacement  volume  of  10 
cubic  feet  and  a  clearance  of  5  per  cent.  What  volume  of  air 
is  taken  in  per  stroke  when  the  compressor  operates  adiabatically 
between  o  and  60  pounds  pe*:  square  inch  gage? 

314.  Sketch  on  the  ^F-plane  the  cycle  thru  which  the 
air  passes  during  the  operation  of  a  dense-air  refrigerating 
machine  under  ideal  conditions.  Indicate  clearly  on  this  diagram 
the  points  at  which  the  air  attains  the  temperature  of  (a)  the 
cooling  water,  (b)  the  cold  storage  room. 

315.  In  a  dense-air  refrigerating  machine  operating  on  a 
reversed  Joule  cycle  the  compressor  takes  in  air  at  a  pressure 
of  40  pounds  per  square  inch  gage  and  delivers  it  at  a  pressure 
of  1 20  pounds  per  square  inch  gage.  At  the  beginning  of  compres- 
sion the  temperature  is  34°  F  and  the  coohng  water  maintains 
the  temperature  of  the  cooling  coils  at  70°  F.  Determine  the 
ratio  of  the  heat  extracted  from  the  cold  body  to  the  heat  equiv- 
alent of  the  work  expended  in  driving  the  machine. 

316.  In  a  dense-air  refrigerating  machine  operating  on  a 
reversed  Joule  cycle,  the  compression  cy Under  takes  in  air  at 
40  pounds  per  square  inch  gage  and  delivers  it  at  120  pounds 


294 


THERMODYNAMICS 


per  square  inch  gage.  At  the  beginning  of  compression  the 
temperature  is  34°  F  and  the  cooling  water  maintains  the  tem- 
perature of  the  cooHng  coils  at  70°  F.  Compute  the  work  expended 
per  pound  of  air  passed  thru  the  cycle.     {R=  53.3.) 

317.  Sketch  the  T^-diagram  corresponding  to  the  />i)-diagram 
shown  in  Fig.  76.  Letter  your  diagram  to  correspond  with 
Fig.  76. 

318.  Compute  the  change  in  entropy  which  occurs  during 
the  process  A  B,  Fig.  76. 

319.  The  clearance  space  of  a  Diesel  engine  is  o.i  of  its  piston 
displacement.      If  the  cylinder  is  full  of  air  at  160°  F  and  14 


pounds  per  square  inch  absolute  at  the  beginning  of  com- 
pression, determine  the  temperature  at  the  end  of  compression 
provided  no  leakage  occurs. 

320.  One  pound  of  air  is  heated  at  constant  pressure  from  60 
to  150"  F.     Compute 

(a)  the  change  in  entropy, 

(b)  the  heat  supplied. 


321.  Prove  that  the  efficiency  of  the  Carnot  cycle  is  i  —  I  ~ 


where  -  is  the  ratio  of  the  least  volume  to  the  volume  before 
r 

adiabatic  compression. 

322.  Air  is   compressed   isothermally   from    70°   F    and    14.7 

pounds    per   square   inch  absolute    to   100  pounds  per  square 


PROBLEMS   FOR   REVIEW 


295 


inch  absolute,  then  expanded    at  constant  pressure,  and  finally 
reduced  to  its  initial  state  along  an  isometric  process. 
Compute  (a)  the  eflSciency  of  this  cycle, 

(b)  the  work  performed  per  pound  of  air. 

323.  In  Fig.  77  is  shown  a  special  case  of  the  Atkinson  cycle. 
This  may  be  considered  as  an  Otto  cycle  in  which  the  expansion 
is  carried  to  the  atmospheric  line.  Compute  the  efficiency  of 
this  cycle  in  terms  of  Ti,  T^,  i\,  Vo,  Vt,  and  the  physical  con- 
stants of  the  gas. 

324.  Compute  the  efficiency  of  the  Lenoir  cycle   (Fig.   78) 


\fr/       ^^p  Atmospheric 

X?«,      ^.^     /Lino 

1 


Fig.  77. 


Fig.  78. 


in  terms  of  the  constant  specific  heats  of  the  gas,  the  temper- 
atures at  I  and  2,  and  the  ratio  of  the  volumes  — . 

Vl 

325.  Sketch  the  Lenoir  cycle  (Fig.  78)  on  the  r5-plane. 
Number  the  important  points  so  as  to  correspond  with  Fig.  78. 

326.  One  pound  of  air  at  80°  F  is  compressed  at  constant 
temperature  from  o  to  150  pounds  per  square  inch  gage.  It 
is  then  expanded  at  constant  pressure  and  finally  expanded 
adiabatically  to  its  initial  state.  Compute  the  highest  tem- 
perature reached  during  the  cycle. 

327.  Compute   the  greatest   and   the   least   volume   occupied 

by  the  air  during  the  cycle  described  in  326. 

T, 

328.  The  efficiency  of  an  Otto  cycle  is   i  — ;p,   where   Ti  is 


296  THERMODYNAMICS 

the  absolute  temperature  at  the  beginning  and  T^  is  the  absolute 
temperature  at  the  end  of  compression. 

The  clearance  space  of  a  6  X  12-inch  Otto  gas  engine  was 
found  to  hold  4  pounds  of  water.  Compute  the  ideal  effi- 
ciency of  this  engine  assuming  adiabatic  expansion  and  com- 
pression. 

329.  The  indicator  card  of  an  Otto  engine  shows  that  the 
pressure  rises  during  compression  from  15  to  200  pounds  per 
square  inch  absolute.  After  ignition  it  is  400  pounds  per  square 
inch  absolute. 

Assuming  the  initial  temperature  of  the  gas  to  be  300°  F 
and  the  compression  to  be  polytropic  {n—  1.3)  what  is  the  probable 
temperature  of  the  gas  after  ignition? 

330.  The  inventor  of  an  oil  engine  claims  for  it  a  consumption 
of  0.31  pound  of  fuel  (yielding  19,000  B.t.u.  per  pound)  per 
indicated  horse-power  hour.  The  edges  of  a  nickel  cube  (melting 
point  2600°  F)  fuse  in  the  cyhnder  of  this  engine  and  the  exhaust 
temperature  is  800°  F.  Test  the  probability  of  the  inventor's 
claim  by  means  of  Carnot's  principle.  State  this  principle  and 
your  conclusion. 

331.  A  room  is  heated  by  means  of  radiators  maintained  at 
220°  F.  The  temperature  of  the  room  is  70°  F,  that  of  the 
outside  air  30°  F.  How  many  B.t.u.  at  70°  F  could  be  trans- 
ferred from  the  outside  air  to  the  room  by  means  of  an  ideal 
heat-motor  combined  with  an  ideal  heat-pump  for  every  100 
B.t.u.  furnished  by  the  radiator? 

332.  (a)  Explain  the  physical  significance  of  a  negative  specific 
heat. 

(6)  For  what  values  of  n  in  the  equation  pv^  =  a,  constant 
is  the  specific  heat  of  a  diatomic  gas  negative  during  the  change 
of  state  described   by  this  equation? 

333.  State  the  second  law  of  thermodynamics  and  prove 
Carnot's  principle. 

334.  Determine  analytically  whether  the  temperature  of  an 


PROBLEMS    FOR   REVIEW  297 

ideal  gas  increases  or  decreases  during  a  polytropic  compression 
for  which  h  =  o.9o. 

335.  A  tank  (capacity  10  cubic  feet)  contains  air  at  atmospheric 
pressure  and  70°  F.  Air  is  forced  into  this  tank  until  the  pres- 
sure rises  to  200  pounds  per  square  inch  gage,  and  the  tem- 
perature to  75°  F.  How  many  pounds  of  air  have  passed  into 
the  tank? 

336.  A  certain  producer  gas  is  said  to  contain  0.6  per  cent  of 
H2,  23.0  per  cent  of  CO,  1.4  per  cent  of  CH4,  n.o  per  cent  of 
CO2,  and  64.0  per  cent  of  X2  by  weight.  Compute  its  gas 
constant. 

337.  Compute  the  mean  specific  heat  of  air  at  constant  volume 
between  60  and  300°  F. 

338.  What  change  in  intrinsic  energy  occurs  when  20  cubic 
feet  of  air  expand  to  30  cubic  feet  under  a  constant  pressure 
of  60  pounds  per  square  inch  absolute? 

339.  (a)  During  an  adiabatic  change  of  state  the  temperature 
of  20  pounds  of  air  changed  from  80  to  150°  F.  How  much 
work  was  done  by  'the  gas  during  this  change? 

(b)  If  the  change  had  been  polytropic  with  11=1.2,  how  much 
work  would  have  been  done  by  the  gas? 

340.  Show  without  the  aid  of  calculus,  that  the  work  re- 
qiiired  to  compress  and  deUver  a  gas  when  the  compression  is 
isothermal  equals  the  work  done  during  compression. 

341.  Thru  what  per  cent  of  the  stroke  must  the  piston  ot 
a  compressor,  having  a  displacement  volume  of  10  cubic  feet 
and  a  clearance  of  5  per  cent,  move  before  the  outlet  valves 
open  when  the  compressor  operates  between  0.5  and  100  pounds 
per  square  inch  gage?     («=  1.3  and  pa=  14.7.) 

342.  The  displacement  volume  of  the  working  piston  of  an 
engine  operating  on  an  Ericsson  cycle  is  to  be  100  cubic  feet, 
the  extreme  pressures  to  be  used  15  and  30  pounds  per  square 
inch  absolute,  the  available  e.xtreme  temperatures  70  and 
500°  F.     (a)  Compute  the  displacement  volume  of  the  compressor 


2g8  THERMODYNAMICS 

piston  neglecting  all  clearances,  (b)  What  horse-power  would 
be  obtained  under  ideal  conditions  if  the  engine  is  single-acting 
and  runs  at  lo  r.p.m.? 

343.  What  is  the  most  important  diflference  between  the 
conditions  existing  during  the  absorption  of  heat  by  the  gas 
during  a  Carnot  and  a  Joule  cycle? 

344.  During  an  Otto  cycle  the  heat  supphed  causes  an  in- 
crease in  pressure  from  60  to  250  pounds  per  square  inch 
absolute.     Compute  the  change  in  entropy  per  pound  of  air. 

345.  Air  contained  in  a  receiver  is  withdrawn  until  the  gage 
shows  28.9  inches  of  vacuum.  If  the  receiver  originally  contained 
air  at  atmospheric  pressure  and  the  barometer  read  30.0  inches, 
what  per  cent  by  weight  of  the  air  originally  present  has  been 
removed? 

346.  (a)  Define  total  heat  of  vapors. 

(b)  Write  the  defining  equation  of  the  heat  content  of 
vapors. 

(c)  Develop  the  relation  between  the  total  heat  and  the  heat 
content  of  a  vapor.     Explain  each  symbol. 

(d)  Which  of  these  values  are  given  in  your  steam  tables? 

347.  (a)  Compute  the  internal  energy  of  steam,  quality 
0.90,  at  100  pounds  per  square  inch  absolute. 

(b)  What  does  the  result  represent? 

(c)  What  is  the  mean  specific  heat  of  superheated  steam  at 
a  constant  pressure  of  150  pounds  per  square  inch  absolute 
between  100  and  300°  F  of  superheat? 

348.  Compute  by  means  of  the  steam  tables  the  quality  of 
steam  after  reversible  adiabatic  expansion  from  200  pounds 
per  square  inch  absolute  and  100°  F  superheat  to  10  pounds 
per  square  inch  absolute. 

349.  Fourteen  pounds  of  steam  are  confined  in  a  closed  re- 
ceiver (capacity  20  cubic  feet).  The  initial  temperature  is  400° 
F.  The  temperature  then  falls  to  70°  F.  (a)  What  is  the  greatest 
and  the  least  pressure  to  which  the  receiver  is  subjected  if  the 


PROBLEMS   FOR   REVIEW  299 

atmospheric  pressure  is  14.5  pounds  per  square  inch?     (b)  Com- 
pute the  heat  lost  by  the  steam. 

350.  One  pound  of  steam  occupies  1.5  cubic  feet.  Compute 
its  entropy,  the  pressure  being  250  pounds  per  square  inch 
absolute? 

351.  One  pound  of  dry  saturated  steam  expands  adiabatically 
from  250  to  15  pounds  per  square  inch  absolute.  Find  the 
external  work  performed  measured  in  B.t.u.  (Use  the  steam 
diagrams.) 

352.  Find  the  volume  of  3  pounds  of  steam  at  100  pounds 
per  square  inch  absolute  and  402°  F. 

353.  The  volume  of  an  engine  cylinder  is  5  cubic  feet.  How 
many  pounds  of  steam,  quality  0.80  at  225  pounds  per  square 
inch  absolute,  are  required  to  fill  it? 

354.  How  much  heat  must  be  extracted  per  pound  of  steam, 
initial  quality  0.90,  confined  in  a  closed  receiver  in  order  to 
reduce  the  pressure  from  120  to  15  pounds  per  square  inch 
absolute? 

355.  Assxmiing  that  steam  expands  adiabatically  in  a  frictionless 
nozzle  from  125  pounds  per  square  inch  absolute,  quality  0.95, 
to  a  temperature  of  40°  F,  what  would  be  the  final  quality? 

356.  A  nozzle  is  to  deliver  5  horse-power  in  the  form  of  kinetic 
energy.  It  is  supplied  with  steam  at  125  pounds  per  square  inch 
absolute,  quality  0.93,  and  discharges  at  a  pressure  of  0.60  pounds 
per  square  inch  absolute.  Compute  the  area  at  the  throat 
and  at  exit,  assuming  the  critical  ratio  to  be  0.579  and  neglecting 
all  frictional  resistances. 

357.  Compute  the  mean  specific  heat  of  steam  at  a  constant 
pressure  of  103  pounds  per  square  inch  absolute  between  440 
and  500°  F. 

358.  Steam  is  cooled  at  constant  volume  from  150°  F  super- 
heat at  160  pounds  per  square  inch  absolute  until  the  pressure 
drops  to  15  pounds  per  square  inch  absolute.  How  much  heat 
must  be  removed  per  pound  of  steam? 


300 


THERMODYNAMICS 


359.  Express  ii—i2,  for  a  reversible  adiabatic  flow  of  steam 
(initial  quality  Xi)  during  which  the  pressure  drops  from  pi 
to  p2,  in  terms  of  X\,  pi,  p2,  and  quantities  which  may  be  found 
in  the  steam  tables.     Use  a  T^-diagram. 

360.  In  a  cycle  steam  changes  its  state  at  constant  volume 
from  50  pounds  per  square  inch  absolute,  quality  0.80,  to  one 
pound  per  square  inch  absolute.  How  much  heat  must  be 
added  per  pound  of  steam  during  this  process? 

361.  Sketch  a  Carnot  and  a  Rankine  cycle  for  steam  on  the 
r^-plane.  In  both  cases  the  expanding  steam  is  to  change  its 
state  from  superheat  to  wet. 

362.  Develop,  by  means  of  the  T^-diagram,  a  formula  from 
which  the  heat  converted  into  work  per  pound  of  wet  steam 
during  a  Rankine  cycle  with  complete  expansion  may  be  com- 
puted by  means  of  the  steam  tables. 

363.  Steam  initially  at  160  pounds  per  square  inch  gage 
and  500°  F  leaves  a  De  Laval  nozzle  at  one  pound  per  square 
inch  gage.  What  is  the  quality  of  the  issuing  steam  if  10  per 
cent  of  the  available  energy  is  lost  in  friction? 

364.  (o)  How  much  work  can  be  obtained  from  one  pound  of 
steam  passing  thru  a  Rankine  cycle  between  pressures  of  140 
and  0.50  pound  per  square  inch  absolute,  provided  the  steam 
is  superheated  250°  F? 

(b)  How  many  pounds  of  steam  must  be  circulated  under 
the  above  conditions  per  i.h.p.  hour? 

365.  Sketch  the  cycle  described  in  364  on  the  Ts-  and  oa 
the  w-planes. 

366.  An  engine  during  a  test  developed  an  indicated  horse- 
power of  170  with  a  boiler  pressure  of  130  pounds  per  square 
inch  absolute,  a  condenser  pressure  of  2  pounds  per  square  inch 
absolute,  and  a  supply  of  2500  pounds  of  dry  saturated  steam 
per  hour.  Compute  the  cylinder  efficiency  and  the  actual 
thermal  efficiency  of  this  engine. 

367.  A  test  on  a  compound  steam  engine  shows  that  386  B.t.u. 


PROBLEMS   FOR    REVIEW 


301 


are  supplied  by  the  boiler  per  i.h.p.  per  minute  when  the  total 
i.h.p.  is  104.6,  the  boiler  pressure  102  pounds  gage,  the  vacuum 
24.4  inches,  the  barometer  29.2  inches,  and  the  priming  i.oo  per 
cent.     Find  the  ideal  and  the  actual  thermal  efficiencies. 

368.  The  test  of  a  non-condensing,  high-speed  engine  showed 
the  i.h.p.  to  be  130,  the  b.h.p.  120,  the  steam  pressure  11 5.3 
pounds  gage,  the  back  pressure  0.30  pound  gage,  the  atmospheric 
pressure  14.7  pounds,  the  quality  of  the  steam  i.oo,  the  steam 
per  i.h.p.  hour  30.5  pounds.     Compute 

(c)  the  ideal  Carnot  efficiency, 

(b)  the  ideal  Rankine  efficiency, 

(c)  the  actual  thermal  efficiency, 
{d)  the  cylinder  efficiency 

under  these  conditions. 

369.  The  test  of  an  engine  shows  that  14  pounds  of  steam 
are  consumed  per  indicated  horse-power  hour.  The  pressure 
and  the  temperature  in  the  steam  main  aie  150  pounds  gage 
and  450°  F.  The  vacuum  maintained  in  the  condenser  was 
28  inches,  while  the  barometer  read  30  inches.  Compute  (a) 
the  actual  thermal  efficiency  of  this  engine,  (6)  the  thermal  effi- 
ciency of  a  Rankine  cycle  with  complete  expansion  operating 
vmder  the  same  conditions. 

370.  Develop  the  two  fundamental  equations  by  means  of 
which  problems  in  the  flow  of  fluids  may  be  solved. 

371.  Compute  the  critical  pressure  for  the  flow  of  a  fluid 
thru  an  orifice  from 


I  2_  TO+l 


IF-. 


where  the  letters  have  their  usual  significance. 

372.  Steam,  quality  0.93,  e.xpands  from  a  pressure  of  iod 
pounds  per  square  inch  absolute  to  0.40  pound  per  square  inch 
absolute  in  a  nozzle.     If    10  per  cent  of  the  heat  transformed 


302  THERMODYNAMICS 

into  kinetic  energy  under  ideal  conditions  is  required  to  over- 
come the  frictional  resistances,  what  would  be  (a)  the  velocity, 
(b)  the  quality  of  the  issuing  steam? 

373.  A  De  Laval  turbine  rated  at  350  horse-power  has  seven 
nozzles.  The  steam  supplied  to  this  turbine  at  180  pounds 
per  square  inch  absolute  is  superheated  70°  F.  The  pressure  of 
the  steam  leaving  the  nozzles  is  one  pound  per  square  inch 
absolute.  The  loss  of  energy  due  to  friction  in  the  nozzles  is 
12  per  cent  and  70  per  cent  of  the  kinetic  energy  developed 
by  the  nozzles  is  available  at  the  brake.     Find 

(a)  the  velocity  of  the  steam  leaving  the  nozzles, 

(b)  the  quality  of  the  steam  leaving  the  nozzles, 

(c)  the  steam  required  per  brake  horse-power  per  hour. 

374.  How  many  pounds  of  dry  saturated  steam  pass  per 
second  thru  a  Venturi  meter  when  pressures  of  100  and  90 
pounds  per  square  inch  absolute  exist  at  the  sections  whose 
areas  are  3  and  2  square  inches  respectively? 

375.  Steam  before  passing  thru  a  throttling  calorimeter  was 
under  a  pressure  of  150  pounds  per  square  inch  gage.  On  the 
down-stream  side  of  the  orifice  the  pressure  is  10  pounds  per 
square  inch  gage,  and  the  temperature  260°  F.  What  was 
the  initial  quality  of  the  steam? 

376.  A  throttling  calorimeter  shows  a  temperature  of  287° 
F  at  16  pounds  per  square  inch  gage.  The  initial  temper- 
ature of  the  steam  in  the  main  was  382°  F,  the  atmospheric 
pressure  being  14.6.  What  was  the  quality  of  the  steam  in 
the  main? 

377.  A  superheat  of  at  least  10°  F  must  be  recorded  when 
a  reliable  determination  of  the  quahty  of  steam  is  to  be  made 
with  a  throtthng  calorimeter.  What  is  the  greatest  percentage 
of  moisture  which  can  be  measured  in  steam  at  200  pounds  per 
square  inch  absolute  when  the  steam  is  throttled  to  15  pounds 
per  square  inch  absolute? 


PROBLEMS   FOR   REVIEW  303 

Indicate  your  method  of  solution  on  a  sketch  of  a  Mollier 
diagram. 

378.  How  much  dry  saturated  steam  will  escape  thru  an 
orifice  (area  |  square  inch)  into  the  atmosphere  (14.7  pounds) 
from  a  receiver  in  which  a  pressure  of  30  pounds  per  square 
inch  absolute  is  maintained.  Solve  by  means  of  the  steam 
diagrams  only. 

379.  Sketch  a  Rankine  cycle  with  incomplete  expansion  on 
(c)  the  pv-plane,  (b)  the  T^-plane,  and  (c)  the  total  heat  entropy 
diagram. 

380.  Sketch  on  the  w-plane  three  adiabatics  between  the 
same  pressures  and  with  the  same  initial  point,  one  for  friction- 
less  flow  thru  a  nozzle,  another  for  flow  thru  a  nozzle  including 
friction,  and  the  third  for  throttling  without  appreciable  change 
of  velocity. 

381.  Compute  the  heat  content  of  one  pound  of  steam  con- 
taining 3  per  cent  of  moisture  at  150  pounds  per  square  inch 
absolute  in  two  different  ways  by  means  of  the  steam  tables. 

382.  It   is  said   that    in   the   notation   and   according   to   the 

formula  of  Exercise  250  the  velocity  of  the  fluid  entering  the 

pipe  is 

„     kpi-p2)dvi    .    ^  , 

Wi  =  C    /  -^. — - — r-,  feet  per  second. 


Show  that  this  is  so,  and  And  the  value  of  C. 


ANSWERS 


Marks  and  Davis  Steam   Tables  and  Diagrams  have  been  used. 


2. 

4. 

7. 

8. 

9. 
11. 
12. 
14. 
15. 

16. 

17. 
21. 
23. 
25. 
29. 

38. 
43. 

47. 
52. 
53. 
57. 


53.3.  3.  204°  F. 

1280  lbs.  per  sq.  in.  abso.  6.  54.9. 

238°  F. 

(a)  86.4  lbs.  per  sq.  in.  abso.;  (h)  76.7  lbs. 

24  lbs. 

(a)  53.2;  (b)  6.4  and  23.6  ins.  of  Hg. 

28.9.  13.  0.0806  lb.  per  cu.  ft. 

(a)  51.1;  (b)  30.1;  (c)  0.0842  lb.  per  cu.  ft. 

(a)  80°  F.;   (fo)  4.4%. 


4.^[-:- 


(a)  4.914,  6.904,  1.406;  (6)  5.020,  7.010,  1.-397. 

777.  22.  (a)  24.0  B.t.u.;  (b)  17.1  B.t.u. 

(a)  183°  F;  (6)  5.75  B.t.u.      24.  389  B.t.u. 

(a)  36  B.t.u.;  (6)  25.6  B.t.u.;  (c)  10.4  B.t.u. 

29,300.  33.  218  B.t.u. 

144  mct,vi{p2  —  pi) 


R 


39.  52.5;  1.016  cu.  ft. 


(a)  188°  F;  (6)   +76.3  B.t.u.;  (c)  191  B.t.u.;  {d)  267  B.t.u. 

t-i 

(a)  ?Jf*  =  piri';  (6)  7^7  =  (~ 

(a)   -180°  F;  (6)  0.814  cu.  ft.;  (c)  6.94  cu.  ft. 

(o)  731°   F;  {b)  —  =18.3;  (c)  883  lbs.  per  sq.  in.  abso. 

Pi 
50,400  ft.-lbs.  58.   -89,500  ft.-lbs. 


3o6  THERMODYNAMICS 

T,     \vj  7\     \]h 

75.  (o)   -83,700  ft.-lbs.;  (b)   +83,700  ft.-lbs. 

76.  (a)   +7460  ft.-lbs.;  (b)   -38.4  B.t.u.;  (c)    -48.0  B.t.u. 

77.  -116,000  ft.-lbs. 

91.  (a)  187  horse-power;  (b)  170  horse-power. 

„,    ,1    P3F3  — 

92.  (a)  J 


-p-iVi  .  P2V2      P3V3]  cv  n-k 

—  1  k  —  l      k  —  l\  Kn  —  \ 


both  of  these  expressions 

m        n  —  k 

-P2V2). 

reduce  to                •          -{pzVs- 
J{k  —  1)  n  — 1 

93.  2.5  lbs.  per  sec;  17  h.p. 

94.  4.16  cu.  ft. 

1 

96.  294  lbs.  per  sq.  in.  abso. 

97    1+c-c     -      . 
\Pi  / 

n-l 

2nNa.p.c.)pJ  /p2\    n   _   1 
■  33,000(n-l)L\Pi/  J 


103.  30.9" X45";  19.1" X45";  417  horse-power. 
105.  680  horse-power.  106.  pipa  =  PiVi' 

107.  163  lbs.  per  sq.  in.  abso. 

RiT,-Tc)lo^.r  ,^^^^^J.^P^^ 

RTHl0ger  +  JCv{TH-Tc)  V2      Pi 

11=    /  ^  'Th-Tc     ,,,  R{TH-Tc)\oger 

115.  (a)        ^^      '  ^'^  flT^log,r+Jc,(r,,-Tc)' 

I'l     r4     Pi      ps 
where  r  =  —  =  —  =  —  = — . 

1'2         1'3        Pi  Pi 

117.  JmcpiTi-To-Ti  +  Ti).  119.  0.77. 

121.  16.8  h.p.;  4  cu.  ft.;  2.74  cu.  ft.     129.  2545  -. 

V 

130.  54%. 

131.  Claimed  effi.,  56.4%;  Carnot  efH.,  57.5%;  No. 


ANSWERS  307 

134.  49.5%.  135.  (a)  35.7%;  (b)  81.8%. 

136.  75.7%.  140.  (a)  0.109;  (b)  0.109. 

141.  -loge— .  143.  0.188. 

J  P2 

150.  2.85  ins.;   -00.        155.  (a)  0.01602  cu.  ft.;  (6)  0.0187  cu.  ft. 
157.  (a)  85.9;  (6)  It  becomes  more  than  ideal. 

160.  -0.03;  +0.18;  0.00;   -2.5;   -7.2  B.t.u. 

161.  0.167  B.t.u.;  ,392.7  B.t.u.       162.  0.844  B.t.u.;  1204.1  B.t.u. 

163.  (o)  1201. 3 -.732 +  .044  =  1200.6  B.t.u.;  (b)  1115.7  B.t.u. 

164.  1.404+0.0047  =  1.409  cu.  ft. 

166.  (a)  115.2  B.t.u.;  (6)  0.576.  167.  231.0  B.t.u. 
176.  (a)  776.5  B.t.u.,  1.045;  (6)  587.1  B.t.u.,  1.0541. 
179.  (a)  0.52;  (6)  185.5  B.t.u.;  (c)  167.0  B.t.u.;  (d)  18.5  B.t.u. 

181.  0.0058. 

182.  (a)  2.062  cu.  ft.;  (b)   223  lbs.  per  sq.  in.  absc,  391°  F. 

183.  -965  B.t.u.  184.  (a)  1.515. 

185.  (a)  0.842;  (fe)  0.308.  187.  1.20  cu.  ft.;  6.83  cu".  ft. 

188.  317  B.t.u.  or  246,000  ft.-lbs. 

192.  (a)  118.5  B.t.u.;  {b)   0.542.  193.  88.5  B.t.u. 

195.  (a)  200.2°  F;  (b)   12.01  cu.  ft.;  (c)  223.8  B.t.u.;  (d)   18.5 

B.t.u.;  (e)  205.3  B.t.u. 

196.  -59.8  B.t.u.  197.  (a)  62.46°  F;  (6)  51.5  B.t.u. 
198.  55  B.t.u.  199.  (a)  0.793;  (b)  327  B.t.u. 
200.  68  lbs.  per  sq.  in.  absc;  71  lbs.  per  sq.  in.  ab.so. 

203.  (a)  0.178;  (6)  0.314;  (r)  0.334. 

204.  77=0.265;  178,000  ft.-lbs.;  11.1  lbs.  per  h.p.-hr. 

206.  (0)  0.286;  (b)  250,000  ft.-lbs.;  (c)  7.9  lbs.  per  i.h.p.-hr. 

207.  (a)  0.318;  (6)  212,500  ft.-lbs.;  (c)  9.33  lbs.  per  i.h.p.-hr. 

208.  (a)  0.294;  (6)  282,500  ft.-lbs.;  (c)  7.01  lbs.  per  i.h.p.-hr. 
219.  (a)  0.189;  (&)  166,000  ft.-lbs.;  (c)  11.98  lbs.  per  i.h.p.-hr. 

212.  (a)  49.5%;  {b)  376  B.t.u.  per  i.h.p.  per  min.;  (c)  11.3%. 

213.  (a)  26.9%;  (6)  8.72  and  (c)  16.0  lbs.  per  i.h.p.-hr.;  (d)  54.5%; 
(e)  14.7%. 


3o8  THERMODYNAMICS 

214.  25  lbs.  per  b.h.p.-hr.  217.  .3878  ft.  per  sec. 

224.  (a)  1491  ft.  per  sec;  (b)   1489  ft.  per  sec. 

225.  2910  ft.  per  sec. 

227.  (a)  38.6;  (6)  38.2;  (c)  38.1  lbs.  per  min. 

228.  0.00606  sq.  in.;  0.188  sq.  in.   230.  204  B.t.u. 

231.  aT^J  =j{h-Z2  +  (s2"-s,)7^2|. 

yiii—ii) 

233.  Si'+XiSi"=S2'+X2S2   ;  Xi=^X2-\ — — 

100r2 

235.  0.0925  in.;  0.535  in.     The  corresponding  values  for  Exercise 

228  are  0.0879  in.,  0.489  in. 

238.  +7.5°  F.  242.  5%. 

243.  0.967.  244.  xi  =  - — -. 

n 

247.  14.4  B.t.u.  249.  10.1  lbs.  per  sec. 

251.  93.2  lbs.  per  min.  252.  24  lbs.  per  sq.  in. 

253.  5.2  ins.;  24  ft.  per  sec.  254.  (a)  3.3  ins.;  (6)  5.6  ins. 

258.  (a)  71.8;  (6)  73.5  lbs.  per  sq.  in.  gage;  (c)  24.6,  30.3  ft.  per  sec. 

263.  (1),  (a)  0,  (6)   -1;  (2),  (a)  3,  (6)  3. 

5  R  RT 

264.  -.  270.  cp-cc=- 


2  J  in-l)Cp''+RT 

pV 
277.   -459.6°  F.  278.  ^. 

oo«     137.5 
281.  149  lbs.  per  sq.  in.  gage.       282.    — — . 

283.  53.3;  53.7.  285.  255°  F. 

290.  247°  F.  291.  6.8  B.t.u. 

292.  58,000  ft.-lbs.  295.  4.86  B.t.u. 

296.   -6.84  B.t.u.  297.  -21,300  ft.-lbs. 

298.  203  lbs. 

299.  a  log,  Uf    +^(^2  -  ^i)  =  -7  ^oge  -. 

\  /  1/  J  Vi 


ANSWERS  309 

300.  418  B.t.u.  301.  43.4  B.t.u. 

302.  920  B.t.u.;  1.53.  304.   -165°  F. 

305.  16.8  B.t.u.  308.  3.54  cu.  ft.;  .390"  F. 

t-i 

310.  nibs.;  1794  lbs.  per  sq.  in.     311.  - — ^J  (-)   "   -1 


k-l\\pj 


k 

312.  (n-1).  313.  8.9  cu.  ft. 

315.  3.42.  316.  4660  ft.-lbs. 

318.  .0473.  319.  1160°  F. 

320.  0.0381;  21.6  B.t.u.  322.  0.193;  142.5  B.t.u. 

cp{T,-T,) 


323.  1- 


324. 


c.\TA'^Y"-tJ'^^'~' 


Vl/     \  \Vy 


mc,iT2-Ti)-mcv{  -  1  ^  7'2l  -  1   -  T,  \  -mcpT 


■{©-' 


7nc„iT2-Ti) 


326.  620°  F.  327.  1.212  cu.  ft.;  13.6  cu.  ft. 

328.  42.9%.  329.  2306°  F. 

330.  43.2%;  Carnoteffi.,  58.8 9o-  331.  371  B.t.u. 

335.  10.1  lbs.  336.  57.7. 

337.  0.173.  338.  277  B.t.u. 

339.   -186,000,-373,000  ft.-lbs.  341.  82.7%. 

342.  55.2  cu.  ft.,  20.3.  344.  0.242. 

345.  96.3%.  347.  (a)  1023.7  B.t.u.;  (c)  0.496. 

348.  0.882. 

349.  233,  14.1  lbs.  per  sq.  in.  gage;  12,600  B.t.u. 

350.  1.3436.  351.  204  B.t.u. 
352.  14.88  cu.  ft.  353.  3.05  lbs. 
354.  734  B.t.u.  355.  0.707. 
356.  0.00606,  0.188  sq.  in.  357.  0.4965. 

358.  867  B.t.u.  360.  -836.2  B.t.u. 

363.  0.941.  364.  400  B.t.u.,  6.36  lbs. 


3IO 


THERMODYNAMICS 


366.  63.5%;  15.8%.  367.  46.5%;  10.99%. 

368.  16.7%;  15.8%;  8.28%;  52.0%.      369.  15.5%;  29.2%. 

372.  3750  ft.  per  sec;  0.762. 

373.  3900  ft.  per  sec;  0.837;  12  lbs. 

374.  2.4  lbs.  per  sec  375.  0.97. 
376.  0.98.  377.  5%. 
378.  0.218  lb.  per  sec. 

381.  330.2+837.2;  1193.4-25.9.  382.  256. 


INDEX 


Absolute  scale,  Kelvin's,  123 
Absolute,  temperature,  13,  123 

zero,  12 
Adiabatic,  change,   38,    144^    i8c 
198 
flow,  218,  239 
Answers,  305 
Availability,  of  heat  energy,  1 28 

loss  of,  248 
Avogadro's  law,  16 

B 

Boyle's  law,  11 
Brayton  cycle,  109 
British  thermal  unit,  4 


Carnot  cycle,  86,  200 
Carnot's  principle,  120 
Changes  of  state,  ^i,  160,  184 
Characteristic,  equations,  30,  155 

surfaces,  31,  153 
Charles'  law,  11 
Clapeyron's  equation,  284 
Clausius'  relation,  286 
Clearance,  75 

Compound  compressors,  77 
Compressors,  71 
Conservation  of  energy,  3,  115 


Continuity,  equation  of,  214 

Critical,  pressure,  157,  223 
states,  157 
temperature,  157 

Cross-products,  100 

Cycles,  gas,  83,  147 
vapor,  200 

Cylinder  efficiency,  211 

D 

D'Arcy's  formula,  258 

Dalton's  law,  18 

Degradation  of  heat  energy,  128 

Diesel  cycle,  112 

Differential  equations,  264 

Discharge,  thru  a  nozzle,  237 

thru  an  orifice,  225,  229 
thru  a  pipe,  258,  262 

Dry  saturated  vapor,  152,  160 


Efficiency,  86,  202,  209 

Energy,  i 

conservation  of,  3 
equation  of,  4,  216 
internal,  2,  65,  66 

Engine  efficiencies,  209 

Entropy,  131,  174 

Entropy  as  a  coordinate,  135,  275 

Ericsson  cycle,  93 


3" 


312 


INDEX 


Exact  differential,  271 
Experimental  determination  of  w. 

External  work,  3,  64 


First  law  of  thermodynamics,  5, 

116 
Fliegner's  formula,  225 
Flow  of  fluids,  213 

thru  a  nozzle,  230 

thru  an  orifice,  221 

thru  pipes,  256 

thru  Venturi  meters,  253 
Friction,  217,  239,  243,  257 
Frictionless  flow,  218 


Gas  constant,  15,  16 

of  mixtures,  18 
Gas  cycles,  83,  147 
Gases,  9 

General  law  for  ideal  gases,  13 
Grashof  s  formula,  227 

H 

Heat,  content,  165,  168,  171 

pump,  102 

supplied,  68,  264,  2S3 
Hydraulic  formula,  256 


Ideal  gas,  9 

Internal  energy,   2,  65,  165,   168, 

172 
Irreversible  processes,  119 
Isodynamic  change,  37 
Isometric  change,  36, 143,  187,  197 
Isopiestic  change,  37,  140,  185,  195 


Isothermal  change,  34,   144,   185, 

196 
Isothermals,  34,  64,  156 


Joule  cycle,  94 
Joule's,  experiment,  10 

law,  10 
Joule-Thomson  coefficient,  287 

K 

Kelvin's    scale   of    temperature, 
123 

L 

Latent  heat,  163 
Liquefaction  of  gases,  157,  246 
Logarithm      temperature-entropy 

diagram,  145 
Loss  of  availability,  248 

M 
Maxwell's  relations,  280 
Meters,  Venturi,  253 
Mollier  diagram,  182,  242 

N 
Napier's  formula,  228 
Nozzles,  230,  236,  243 


Orifice,  flow  thru  an,  221 
Otto  cycle,  109 

P 
Perfect  gas,  9     « 

Perpetual  motion,  129 
Pipes,  flow  thru,  256 
Polytropic,  change,  45,  144 

plotting,  61 
Potentials,  278 
Problems  for  review,  289 


INDEX 


313 


Quality,  167,  168 


Rankine  cycle,  203,  207 
Refrigerating  machine,  102 
Reversible  processes,  117 
Review,  problems  for,  289 


Saint  Venant,  223 

Second   law   of   thermodynamics, 
116,  122 

Single-stage  compressors,  71 

Specific  heat,  6 

of  gases,  20 

at  constant  volume,  21 

at  constant  pressure,  23,  26 

Specific  heats,  negative,  52 

Stirling  cycle,  91 


Temperature-entropy  diagram, 

140,  145,  176,  182,  237 
Thermal,  capacity,  5,  267 

efficiency,  86,  211 

unit,  4 
Thermodynamic  potentials,  278 
Three-stage  compressors,  80 
Throttling,  244 
Total  heat,  162,  164,  168 
Transformation  of  energy,  115 
Turbo-compressors,  81 

V 

Van  der  Waals'  equation,  158 

Vapor  cycles,  200 

Vapors,  dry  saturated,  152 
wet,  167,  184 
superheated,  169,  195 

Venturi  meters,  253 

Volumetric  efficiency,  77 

Von  Linde,  246 

W 
Warming  engine,  106 
Wet  vapors,  167 


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